Разложите на множители 2a в пятой степени - 2a в 3 степени(-2a в 3 степени)рядом с 2 минус а не тире
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Ответ: 4)
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Ответ: 2)
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![\sqrt{x-2}=8-x\\\\ \left \{ {{x-2 \geq 0} \atop {8-x \geq 0}} \right. \\\\ \left \{ {{x \geq 2} \atop {-x \geq -8}} \right. \\\\ \left \{ {{x \geq 2} \atop {x \leq 8}} \right. \\\\ x\in [2;8]\\\\ ( \sqrt{x-2})^2=(8-x)^2\\ x-2=64-16x+x^2\\ 64-16x+x^2-x+2=0\\ x^2-17x+66=0\\ D=289-264=25\\ x_1= \frac{17+5}{2} = \frac{22}{2}=11 \notin [2;8]\\ x_2= \frac{17-5}{2} = \frac{12}{2}=6\\ ](https://tex.z-dn.net/?f=+%5Csqrt%7Bx-2%7D%3D8-x%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx-2+%5Cgeq+0%7D+%5Catop+%7B8-x+%5Cgeq+0%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+2%7D+%5Catop+%7B-x+%5Cgeq+-8%7D%7D+%5Cright.+%5C%5C%5C%5C%0A+%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+2%7D+%5Catop+%7Bx++%5Cleq++8%7D%7D+%5Cright.+%5C%5C%5C%5C%0Ax%5Cin+%5B2%3B8%5D%5C%5C%5C%5C%0A+%28+%5Csqrt%7Bx-2%7D%29%5E2%3D%288-x%29%5E2%5C%5C%0Ax-2%3D64-16x%2Bx%5E2%5C%5C%0A64-16x%2Bx%5E2-x%2B2%3D0%5C%5C%0Ax%5E2-17x%2B66%3D0%5C%5C%0AD%3D289-264%3D25%5C%5C%0Ax_1%3D+%5Cfrac%7B17%2B5%7D%7B2%7D+%3D+%5Cfrac%7B22%7D%7B2%7D%3D11+%5Cnotin+%5B2%3B8%5D%5C%5C%0Ax_2%3D+%5Cfrac%7B17-5%7D%7B2%7D+%3D+%5Cfrac%7B12%7D%7B2%7D%3D6%5C%5C%0A+%0A%0A%0A+)
Ответ 3)
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Ответ: 4)
Ответ: 4)
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Ответ: 2)
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Ответ 3)
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Ответ: 4)