miniaufgabe.js d3.min.js function-plot.js ==== 26. September 2022 bis 30. September 2022 ==== === Dienstag 27. September 2022 === Lösen sie das folgende Gleichungssystem auf:miniAufgabe("#exolinglsyst_2","#sollinglsyst_2", [["$$\\left\\{\\begin{array}{rcrcrcrr}\n3x&- & 3y&+ & z&= & 1 & (G_{0})\\\\\n3x&+ & 2y&- & 3z&= & -1 & (G_{1})\\\\\n-3x&+ & 7y&- & 5z&= & -1 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n3x&- & 3y&+ & z&= & 1 & (G_{0})\\\\\n3x&+ & 2y&- & 3z&= & -1 & (G_{1})\\\\\n-3x&+ & 7y&- & 5z&= & -1 & (G_{2})\\end{array}\\right.$$\nVariable $x$ eliminieren:\n\\begin{align*}\n(G_{0})-(G_{1}): && -5 & y &+4 & z & = 2 && (G_{0}')\\\\\n(G_{1})+(G_{2}): && 9 & y &-8 & z & = -2 && (G_{1}')\\\\\n\\end{align*}\nVariable $z$ eliminieren:\n\\begin{align*}\n2(G_{0}')+(G_{1}'): && - & y & = 2 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $y = -2$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n-5 \\cdot (-2)+4z & = 2 && |\\text{TU}\\\\\n4z+10 & = 2 &&|-10\\\\\n4z & = -8 &&|:4\\\\\nz & = -2 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\n3x-3 \\cdot (-2)+ \\cdot (-2) & = 1 && |\\text{TU}\\\\\n3x+4 & = 1 &&|-4\\\\\n3x & = -3 &&|:3\\\\\nx & = -1 \\\\\n\\end{align*}\n\nLösung: $x = -1$, $y = -2$, $z = -2$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n-9x&- & 5y&- & z&= & -2 & (G_{0})\\\\\n-8x&- & 2y&+ & 2z&= & -4 & (G_{1})\\\\\n5x&+ & 3y&+ & 2z&= & -5 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n-9x&- & 5y&- & z&= & -2 & (G_{0})\\\\\n-8x&- & 2y&+ & 2z&= & -4 & (G_{1})\\\\\n5x&+ & 3y&+ & 2z&= & -5 & (G_{2})\\end{array}\\right.$$\nVariable $z$ eliminieren:\n\\begin{align*}\n2(G_{0})+(G_{1}): && -26 & x &-12 & y & = -8 && (G_{0}')\\\\\n(G_{1})-(G_{2}): && -13 & x &-5 & y & = 1 && (G_{1}')\\\\\n\\end{align*}\nVariable $x$ eliminieren:\n\\begin{align*}\n(G_{0}')-2(G_{1}'): && -2 & y & = -10 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $y = 5$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n-26x-12 \\cdot 5 & = -8 && |\\text{TU}\\\\\n-26x-60 & = -8 &&|+60\\\\\n-26x & = 52 &&|:-26\\\\\nx & = -2 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\n-9 \\cdot (-2)-5 \\cdot 5-z & = -2 && |\\text{TU}\\\\\n-z-7 & = -2 &&|+7\\\\\n-z & = 5 &&|:-1\\\\\nz & = -5 \\\\\n\\end{align*}\n\nLösung: $x = -2$, $y = 5$, $z = -5$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\nx&- & y&- & z&= & -1 & (G_{0})\\\\\n-x&+ & 5y&+ & 4z&= & 2 & (G_{1})\\\\\n-x&- & 5y&- & 3z&= & -3 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\nx&- & y&- & z&= & -1 & (G_{0})\\\\\n-x&+ & 5y&+ & 4z&= & 2 & (G_{1})\\\\\n-x&- & 5y&- & 3z&= & -3 & (G_{2})\\end{array}\\right.$$\nVariable $x$ eliminieren:\n\\begin{align*}\n(G_{0})+(G_{1}): && 4 & y &+3 & z & = 1 && (G_{0}')\\\\\n(G_{1})-(G_{2}): && 10 & y &+7 & z & = 5 && (G_{1}')\\\\\n\\end{align*}\nVariable $y$ eliminieren:\n\\begin{align*}\n5(G_{0}')-2(G_{1}'): && & z & = -5 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $z = -5$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n4y+3 \\cdot (-5) & = 1 && |\\text{TU}\\\\\n4y-15 & = 1 &&|+15\\\\\n4y & = 16 &&|:4\\\\\ny & = 4 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\nx- \\cdot 4- \\cdot (-5) & = -1 && |\\text{TU}\\\\\nx+ & = -1 &&|-\\\\\nx & = -2\\\\\n\\end{align*}\n\nLösung: $x = -2$, $y = 4$, $z = -5$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n-2x&+ & y&+ & 3z&= & -1 & (G_{0})\\\\\n-5x&- & y&+ & 5z&= & 4 & (G_{1})\\\\\n-x&- & 8y&- & z&= & 1 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n-2x&+ & y&+ & 3z&= & -1 & (G_{0})\\\\\n-5x&- & y&+ & 5z&= & 4 & (G_{1})\\\\\n-x&- & 8y&- & z&= & 1 & (G_{2})\\end{array}\\right.$$\nVariable $x$ eliminieren:\n\\begin{align*}\n(G_{0})-2(G_{2}): && 17 & y &+5 & z & = -3 && (G_{0}')\\\\\n(G_{1})-5(G_{2}): && 39 & y &+10 & z & = -1 && (G_{1}')\\\\\n\\end{align*}\nVariable $z$ eliminieren:\n\\begin{align*}\n2(G_{0}')-(G_{1}'): && -5 & y & = -5 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $y = 1$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n17 \\cdot 1+5z & = -3 && |\\text{TU}\\\\\n5z+17 & = -3 &&|-17\\\\\n5z & = -20 &&|:5\\\\\nz & = -4 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\n-2x+ \\cdot 1+3 \\cdot (-4) & = -1 && |\\text{TU}\\\\\n-2x-11 & = -1 &&|+11\\\\\n-2x & = 10 &&|:-2\\\\\nx & = -5 \\\\\n\\end{align*}\n\nLösung: $x = -5$, $y = 1$, $z = -4$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n-4x&- & 2y&- & 2z&= & -4 & (G_{0})\\\\\n-5x&+ & 3y&+ & z&= & -1 & (G_{1})\\\\\n7x&- & 4y&- & z&= & 1 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n-4x&- & 2y&- & 2z&= & -4 & (G_{0})\\\\\n-5x&+ & 3y&+ & z&= & -1 & (G_{1})\\\\\n7x&- & 4y&- & z&= & 1 & (G_{2})\\end{array}\\right.$$\nVariable $z$ eliminieren:\n\\begin{align*}\n(G_{0})+2(G_{1}): && -14 & x &+4 & y & = -6 && (G_{0}')\\\\\n(G_{1})+(G_{2}): && 2 & x &- & y & = 0 && (G_{1}')\\\\\n\\end{align*}\nVariable $y$ eliminieren:\n\\begin{align*}\n(G_{0}')+4(G_{1}'): && -6 & x & = -6 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $x = 1$. Eingesetzt in $(G_1')$:\n\\begin{align*}\n2 \\cdot 1-y & = 0 && |\\text{TU}\\\\\n-y+2 & = 0 &&|-2\\\\\n-y & = -2 &&|:-1\\\\\ny & = 2 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\n-4 \\cdot 1-2 \\cdot 2-2z & = -4 && |\\text{TU}\\\\\n-2z-8 & = -4 &&|+8\\\\\n-2z & = 4 &&|:-2\\\\\nz & = -2 \\\\\n\\end{align*}\n\nLösung: $x = 1$, $y = 2$, $z = -2$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\nx&- & 3y&- & 7z&= & -1 & (G_{0})\\\\\n5x&+ & 2y&- & z&= & -5 & (G_{1})\\\\\n-5x&+ & y&+ & 3z&= & -3 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\nx&- & 3y&- & 7z&= & -1 & (G_{0})\\\\\n5x&+ & 2y&- & z&= & -5 & (G_{1})\\\\\n-5x&+ & y&+ & 3z&= & -3 & (G_{2})\\end{array}\\right.$$\nVariable $y$ eliminieren:\n\\begin{align*}\n(G_{0})+3(G_{2}): && -14 & x &+2 & z & = -10 && (G_{0}')\\\\\n(G_{1})-2(G_{2}): && 15 & x &-7 & z & = 1 && (G_{1}')\\\\\n\\end{align*}\nVariable $z$ eliminieren:\n\\begin{align*}\n7(G_{0}')+2(G_{1}'): && -68 & x & = -68 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $x = 1$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n-14 \\cdot 1+2z & = -10 && |\\text{TU}\\\\\n2z-14 & = -10 &&|+14\\\\\n2z & = 4 &&|:2\\\\\nz & = 2 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\n \\cdot 1-3y-7 \\cdot 2 & = -1 && |\\text{TU}\\\\\n-3y-13 & = -1 &&|+13\\\\\n-3y & = 12 &&|:-3\\\\\ny & = -4 \\\\\n\\end{align*}\n\nLösung: $x = 1$, $y = -4$, $z = 2$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n8x&- & 3y&- & 4z&= & -1 & (G_{0})\\\\\n-7x&- & 6y&- & 2z&= & -3 & (G_{1})\\\\\n8x&+ & 9y&+ & 5z&= & -1 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n8x&- & 3y&- & 4z&= & -1 & (G_{0})\\\\\n-7x&- & 6y&- & 2z&= & -3 & (G_{1})\\\\\n8x&+ & 9y&+ & 5z&= & -1 & (G_{2})\\end{array}\\right.$$\nVariable $y$ eliminieren:\n\\begin{align*}\n2(G_{0})-(G_{1}): && 23 & x &-6 & z & = 1 && (G_{0}')\\\\\n3(G_{1})+2(G_{2}): && -5 & x &+4 & z & = -11 && (G_{1}')\\\\\n\\end{align*}\nVariable $z$ eliminieren:\n\\begin{align*}\n2(G_{0}')+3(G_{1}'): && 31 & x & = -31 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $x = -1$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n23 \\cdot (-1)-6z & = 1 && |\\text{TU}\\\\\n-6z-23 & = 1 &&|+23\\\\\n-6z & = 24 &&|:-6\\\\\nz & = -4 \\\\\n\\end{align*}\nEingesetzt in $(G_0)$:\n\\begin{align*}\n8 \\cdot (-1)-3y-4 \\cdot (-4) & = -1 && |\\text{TU}\\\\\n-3y+8 & = -1 &&|-8\\\\\n-3y & = -9 &&|:-3\\\\\ny & = 3 \\\\\n\\end{align*}\n\nLösung: $x = -1$, $y = 3$, $z = -4$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n-2x&+ & 2y&+ & 5z&= & -3 & (G_{0})\\\\\n-5x&+ & y&+ & 9z&= & -2 & (G_{1})\\\\\n-4x&+ & y&+ & 9z&= & 3 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n-2x&+ & 2y&+ & 5z&= & -3 & (G_{0})\\\\\n-5x&+ & y&+ & 9z&= & -2 & (G_{1})\\\\\n-4x&+ & y&+ & 9z&= & 3 & (G_{2})\\end{array}\\right.$$\nVariable $y$ eliminieren:\n\\begin{align*}\n(G_{0})-2(G_{1}): && 8 & x &-13 & z & = 1 && (G_{0}')\\\\\n(G_{1})-(G_{2}): && - & x & && = -5 && (G_{1}')\\\\\n\\end{align*}\nVariable $z$ eliminieren:\n\\begin{align*}\n(G_1'): && - & x & = -5 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $x = 5$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n8 \\cdot 5-13z & = 1 && |\\text{TU}\\\\\n-13z+40 & = 1 &&|-40\\\\\n-13z & = -39 &&|:-13\\\\\nz & = 3 \\\\\n\\end{align*}\nEingesetzt in $(G_1)$:\n\\begin{align*}\n-5 \\cdot 5+y+9 \\cdot 3 & = -2 && |\\text{TU}\\\\\ny+2 & = -2 &&|-2\\\\\ny & = -4\\\\\n\\end{align*}\n\nLösung: $x = 5$, $y = -4$, $z = 3$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n-3x&+ & 2y&- & 6z&= & -2 & (G_{0})\\\\\n-4x&+ & y&- & 4z&= & 1 & (G_{1})\\\\\nx&+ & 3y&- & 4z&= & 1 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n-3x&+ & 2y&- & 6z&= & -2 & (G_{0})\\\\\n-4x&+ & y&- & 4z&= & 1 & (G_{1})\\\\\nx&+ & 3y&- & 4z&= & 1 & (G_{2})\\end{array}\\right.$$\nVariable $y$ eliminieren:\n\\begin{align*}\n(G_{0})-2(G_{1}): && 5 & x &+2 & z & = -4 && (G_{0}')\\\\\n3(G_{1})-(G_{2}): && -13 & x &-8 & z & = 2 && (G_{1}')\\\\\n\\end{align*}\nVariable $z$ eliminieren:\n\\begin{align*}\n4(G_{0}')+(G_{1}'): && 7 & x & = -14 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $x = -2$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n5 \\cdot (-2)+2z & = -4 && |\\text{TU}\\\\\n2z-10 & = -4 &&|+10\\\\\n2z & = 6 &&|:2\\\\\nz & = 3 \\\\\n\\end{align*}\nEingesetzt in $(G_1)$:\n\\begin{align*}\n-4 \\cdot (-2)+y-4 \\cdot 3 & = 1 && |\\text{TU}\\\\\ny-4 & = 1 &&|+4\\\\\ny & = 5\\\\\n\\end{align*}\n\nLösung: $x = -2$, $y = 5$, $z = 3$"], ["$$\\left\\{\\begin{array}{rcrcrcrr}\n2x&- & 3y&- & 5z&= & 1 & (G_{0})\\\\\n-x&+ & 2y&+ & 3z&= & -2 & (G_{1})\\\\\n-2x&- & y&- & 4z&= & 1 & (G_{2})\\end{array}\\right.$$\n", "$$\\left\\{\\begin{array}{rcrcrcrr}\n2x&- & 3y&- & 5z&= & 1 & (G_{0})\\\\\n-x&+ & 2y&+ & 3z&= & -2 & (G_{1})\\\\\n-2x&- & y&- & 4z&= & 1 & (G_{2})\\end{array}\\right.$$\nVariable $x$ eliminieren:\n\\begin{align*}\n(G_{0})+(G_{2}): && -4 & y &-9 & z & = 2 && (G_{0}')\\\\\n2(G_{1})+(G_{0}): && & y &+ & z & = -3 && (G_{1}')\\\\\n\\end{align*}\nVariable $y$ eliminieren:\n\\begin{align*}\n(G_{0}')+4(G_{1}'): && -5 & z & = -10 && (G_{0}'')\\\\\n\\end{align*}\nAus $(G_0'')$ folgt: $z = 2$. Eingesetzt in $(G_0')$:\n\\begin{align*}\n-4y-9 \\cdot 2 & = 2 && |\\text{TU}\\\\\n-4y-18 & = 2 &&|+18\\\\\n-4y & = 20 &&|:-4\\\\\ny & = -5 \\\\\n\\end{align*}\nEingesetzt in $(G_1)$:\n\\begin{align*}\n-x+2 \\cdot (-5)+3 \\cdot 2 & = -2 && |\\text{TU}\\\\\n-x-4 & = -2 &&|+4\\\\\n-x & = 2 &&|:-1\\\\\nx & = -2 \\\\\n\\end{align*}\n\nLösung: $x = -2$, $y = -5$, $z = 2$"]], "
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ruby linearegleichungssystem-mit-linearkombination.rb 2
 === Donnerstag 29. September 2022 === Ausrechnen, Resultat als vollständig gekürzter Bruch. **Achtung:** Potenzen erst ganz am Schluss ausrechnen. Zuerst Basen in Primfaktoren zerlegen und vor dem Multiplizieren kürzen!miniAufgabe("#exonegativeexponenten","#solnegativeexponenten", [["$\\displaystyle \\left(\\frac{3}{2}-\\frac{11}{10}\\right)^{-2} \\cdot \\left(-\\frac{3}{2}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{3}{2}-\\frac{11}{10}\\right)^{-2} \\cdot \\left(-\\frac{3}{2}\\right)^{-3} = \\left(\\frac{15}{10}-\\frac{11}{10}\\right)^{-2} \\cdot \\left(-\\frac{3}{2}\\right)^{-3} = \\left(\\frac{2}{5}\\right)^{-2} \\cdot \\left(-\\frac{2}{3}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(-\\frac{2}{3}\\right)^3 = \\left(\\frac{5}{2}\\right)^2 \\cdot \\left(-\\frac{2}{3}\\right)^3 = \\frac{5^{2}}{2^{2}} \\cdot -\\frac{2^{3}}{3^{3}} = -\\frac{2 \\cdot 5^{2}}{3^{3}} = -\\frac{50}{27}$"], ["$\\displaystyle \\left(\\frac{1}{2}-\\frac{4}{3}\\right)^{-2} \\cdot \\left(\\frac{4}{5}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{1}{2}-\\frac{4}{3}\\right)^{-2} \\cdot \\left(\\frac{4}{5}\\right)^{-3} = \\left(\\frac{3}{6}-\\frac{8}{6}\\right)^{-2} \\cdot \\left(\\frac{4}{5}\\right)^{-3} = \\left(-\\frac{5}{6}\\right)^{-2} \\cdot \\left(\\frac{5}{4}\\right)^3 = \\left(\\frac{6}{5}\\right)^2 \\cdot \\left(\\frac{5}{4}\\right)^3 = \\left(\\frac{2 \\cdot 3}{5}\\right)^2 \\cdot \\left(\\frac{5}{2^{2}}\\right)^3 = \\frac{2^{2} \\cdot 3^{2}}{5^{2}} \\cdot \\frac{5^{3}}{2^{6}} = \\frac{3^{2} \\cdot 5}{2^{4}} = \\frac{45}{16}$"], ["$\\displaystyle \\left(-\\frac{5}{6}+\\frac{4}{3}\\right)^{-2} \\cdot \\left(-\\frac{4}{3}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{5}{6}+\\frac{4}{3}\\right)^{-2} \\cdot \\left(-\\frac{4}{3}\\right)^{-3} = \\left(-\\frac{5}{6}+\\frac{8}{6}\\right)^{-2} \\cdot \\left(-\\frac{4}{3}\\right)^{-3} = \\left(\\frac{1}{2}\\right)^{-2} \\cdot \\left(-\\frac{3}{4}\\right)^3 = \\left(2\\right)^2 \\cdot \\left(-\\frac{3}{4}\\right)^3 = \\left(\\frac{2}{1}\\right)^2 \\cdot \\left(-\\frac{3}{2^{2}}\\right)^3 = \\frac{2^{2}}{1^{2}} \\cdot -\\frac{3^{3}}{2^{6}} = -\\frac{3^{3}}{1^{2} \\cdot 2^{4}} = -\\frac{27}{16}$"], ["$\\displaystyle \\left(\\frac{5}{6}-\\frac{1}{12}\\right)^{-2} \\cdot \\left(-\\frac{4}{9}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{5}{6}-\\frac{1}{12}\\right)^{-2} \\cdot \\left(-\\frac{4}{9}\\right)^{-3} = \\left(\\frac{10}{12}-\\frac{1}{12}\\right)^{-2} \\cdot \\left(-\\frac{4}{9}\\right)^{-3} = \\left(\\frac{3}{4}\\right)^{-2} \\cdot \\left(-\\frac{9}{4}\\right)^3 = \\left(\\frac{4}{3}\\right)^2 \\cdot \\left(-\\frac{9}{4}\\right)^3 = \\left(\\frac{2^{2}}{3}\\right)^2 \\cdot \\left(-\\frac{3^{2}}{2^{2}}\\right)^3 = \\frac{2^{4}}{3^{2}} \\cdot -\\frac{3^{6}}{2^{6}} = -\\frac{3^{4}}{2^{2}} = -\\frac{81}{4}$"], ["$\\displaystyle \\left(\\frac{2}{9}-\\frac{13}{18}\\right)^{-2} \\cdot \\left(\\frac{4}{3}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{2}{9}-\\frac{13}{18}\\right)^{-2} \\cdot \\left(\\frac{4}{3}\\right)^{-3} = \\left(\\frac{4}{18}-\\frac{13}{18}\\right)^{-2} \\cdot \\left(\\frac{4}{3}\\right)^{-3} = \\left(-\\frac{1}{2}\\right)^{-2} \\cdot \\left(\\frac{3}{4}\\right)^3 = \\left(2\\right)^2 \\cdot \\left(\\frac{3}{4}\\right)^3 = \\left(\\frac{2}{1}\\right)^2 \\cdot \\left(\\frac{3}{2^{2}}\\right)^3 = \\frac{2^{2}}{1^{2}} \\cdot \\frac{3^{3}}{2^{6}} = \\frac{3^{3}}{1^{2} \\cdot 2^{4}} = \\frac{27}{16}$"], ["$\\displaystyle \\left(-\\frac{3}{10}+\\frac{7}{90}\\right)^{-2} \\cdot \\left(-\\frac{9}{4}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{3}{10}+\\frac{7}{90}\\right)^{-2} \\cdot \\left(-\\frac{9}{4}\\right)^{-3} = \\left(-\\frac{27}{90}+\\frac{7}{90}\\right)^{-2} \\cdot \\left(-\\frac{9}{4}\\right)^{-3} = \\left(-\\frac{2}{9}\\right)^{-2} \\cdot \\left(-\\frac{4}{9}\\right)^3 = \\left(\\frac{9}{2}\\right)^2 \\cdot \\left(-\\frac{4}{9}\\right)^3 = \\left(\\frac{3^{2}}{2}\\right)^2 \\cdot \\left(-\\frac{2^{2}}{3^{2}}\\right)^3 = \\frac{3^{4}}{2^{2}} \\cdot -\\frac{2^{6}}{3^{6}} = -\\frac{2^{4}}{3^{2}} = -\\frac{16}{9}$"], ["$\\displaystyle \\left(\\frac{1}{2}+\\frac{1}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{6}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{1}{2}+\\frac{1}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{6}\\right)^{-3} = \\left(\\frac{5}{10}+\\frac{1}{10}\\right)^{-2} \\cdot \\left(-\\frac{5}{6}\\right)^{-3} = \\left(\\frac{3}{5}\\right)^{-2} \\cdot \\left(-\\frac{6}{5}\\right)^3 = \\left(\\frac{5}{3}\\right)^2 \\cdot \\left(-\\frac{6}{5}\\right)^3 = \\left(\\frac{5}{3}\\right)^2 \\cdot \\left(-\\frac{2 \\cdot 3}{5}\\right)^3 = \\frac{5^{2}}{3^{2}} \\cdot -\\frac{2^{3} \\cdot 3^{3}}{5^{3}} = -\\frac{2^{3} \\cdot 3}{5} = -\\frac{24}{5}$"], ["$\\displaystyle \\left(-\\frac{3}{8}-\\frac{11}{24}\\right)^{-2} \\cdot \\left(\\frac{4}{5}\\right)^{-3}$", "$\\displaystyle \\left(-\\frac{3}{8}-\\frac{11}{24}\\right)^{-2} \\cdot \\left(\\frac{4}{5}\\right)^{-3} = \\left(-\\frac{9}{24}-\\frac{11}{24}\\right)^{-2} \\cdot \\left(\\frac{4}{5}\\right)^{-3} = \\left(-\\frac{5}{6}\\right)^{-2} \\cdot \\left(\\frac{5}{4}\\right)^3 = \\left(\\frac{6}{5}\\right)^2 \\cdot \\left(\\frac{5}{4}\\right)^3 = \\left(\\frac{2 \\cdot 3}{5}\\right)^2 \\cdot \\left(\\frac{5}{2^{2}}\\right)^3 = \\frac{2^{2} \\cdot 3^{2}}{5^{2}} \\cdot \\frac{5^{3}}{2^{6}} = \\frac{3^{2} \\cdot 5}{2^{4}} = \\frac{45}{16}$"], ["$\\displaystyle \\left(\\frac{1}{3}-\\frac{16}{21}\\right)^{-2} \\cdot \\left(-\\frac{7}{6}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{1}{3}-\\frac{16}{21}\\right)^{-2} \\cdot \\left(-\\frac{7}{6}\\right)^{-3} = \\left(\\frac{7}{21}-\\frac{16}{21}\\right)^{-2} \\cdot \\left(-\\frac{7}{6}\\right)^{-3} = \\left(-\\frac{3}{7}\\right)^{-2} \\cdot \\left(-\\frac{6}{7}\\right)^3 = \\left(\\frac{7}{3}\\right)^2 \\cdot \\left(-\\frac{6}{7}\\right)^3 = \\left(\\frac{7}{3}\\right)^2 \\cdot \\left(-\\frac{2 \\cdot 3}{7}\\right)^3 = \\frac{7^{2}}{3^{2}} \\cdot -\\frac{2^{3} \\cdot 3^{3}}{7^{3}} = -\\frac{2^{3} \\cdot 3}{7} = -\\frac{24}{7}$"], ["$\\displaystyle \\left(\\frac{1}{2}-\\frac{1}{5}\\right)^{-2} \\cdot \\left(\\frac{5}{2}\\right)^{-3}$", "$\\displaystyle \\left(\\frac{1}{2}-\\frac{1}{5}\\right)^{-2} \\cdot \\left(\\frac{5}{2}\\right)^{-3} = \\left(\\frac{5}{10}-\\frac{2}{10}\\right)^{-2} \\cdot \\left(\\frac{5}{2}\\right)^{-3} = \\left(\\frac{3}{10}\\right)^{-2} \\cdot \\left(\\frac{2}{5}\\right)^3 = \\left(\\frac{10}{3}\\right)^2 \\cdot \\left(\\frac{2}{5}\\right)^3 = \\left(\\frac{2 \\cdot 5}{3}\\right)^2 \\cdot \\left(\\frac{2}{5}\\right)^3 = \\frac{2^{2} \\cdot 5^{2}}{3^{2}} \\cdot \\frac{2^{3}}{5^{3}} = \\frac{2^{5}}{3^{2} \\cdot 5} = \\frac{32}{45}$"]], "
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