Пусть . Тогда
1)
Пусть
2)
Все корни проходят проверку x ≠ 1.
Ответ:
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Verified answer
Пусть
. Тогда
1)![\frac{x-1}{4}-\frac{2}{x-1}=3\\ \frac{(x-1)^2-8}{4(x-1)}-3=0\\ \frac{(x-1)^2-12(x-1)-8}{4(x-1)}=0 \frac{x-1}{4}-\frac{2}{x-1}=3\\ \frac{(x-1)^2-8}{4(x-1)}-3=0\\ \frac{(x-1)^2-12(x-1)-8}{4(x-1)}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7B4%7D-%5Cfrac%7B2%7D%7Bx-1%7D%3D3%5C%5C%20%5Cfrac%7B%28x-1%29%5E2-8%7D%7B4%28x-1%29%7D-3%3D0%5C%5C%20%5Cfrac%7B%28x-1%29%5E2-12%28x-1%29-8%7D%7B4%28x-1%29%7D%3D0)
Пусть![x-1=a x-1=a](https://tex.z-dn.net/?f=x-1%3Da)
2)![\frac{x-1}{4}-\frac{2}{x-1}=\frac{1}{2}\\\frac{(x-1)^2-8}{4(x-1)}-\frac{1}{2}=0\\\frac{(x-1)^2-2(x-1)-8}{4(x-1)}=0\\a^2-2a-8=0\\(a-1)^2-9=0\\(a-4)(a+2)=0\\a_{1}=x-1=4\Leftrightarrow x=5\\a_{2}=x-1=-2\Leftrightarrow x=-1 \frac{x-1}{4}-\frac{2}{x-1}=\frac{1}{2}\\\frac{(x-1)^2-8}{4(x-1)}-\frac{1}{2}=0\\\frac{(x-1)^2-2(x-1)-8}{4(x-1)}=0\\a^2-2a-8=0\\(a-1)^2-9=0\\(a-4)(a+2)=0\\a_{1}=x-1=4\Leftrightarrow x=5\\a_{2}=x-1=-2\Leftrightarrow x=-1](https://tex.z-dn.net/?f=%5Cfrac%7Bx-1%7D%7B4%7D-%5Cfrac%7B2%7D%7Bx-1%7D%3D%5Cfrac%7B1%7D%7B2%7D%5C%5C%5Cfrac%7B%28x-1%29%5E2-8%7D%7B4%28x-1%29%7D-%5Cfrac%7B1%7D%7B2%7D%3D0%5C%5C%5Cfrac%7B%28x-1%29%5E2-2%28x-1%29-8%7D%7B4%28x-1%29%7D%3D0%5C%5Ca%5E2-2a-8%3D0%5C%5C%28a-1%29%5E2-9%3D0%5C%5C%28a-4%29%28a%2B2%29%3D0%5C%5Ca_%7B1%7D%3Dx-1%3D4%5CLeftrightarrow%20x%3D5%5C%5Ca_%7B2%7D%3Dx-1%3D-2%5CLeftrightarrow%20x%3D-1)
Все корни проходят проверку x ≠ 1.
Ответ:![-1; 5; 7\pm2\sqrt{11} -1; 5; 7\pm2\sqrt{11}](https://tex.z-dn.net/?f=-1%3B%205%3B%207%5Cpm2%5Csqrt%7B11%7D)