Предмет: Алгебра,
автор: FlexxZ000
Решите уравнение
Только пожалуйста распишите поподробней....![\sqrt{x+1}- \sqrt{x-7}=2 \sqrt{x+1}- \sqrt{x-7}=2](https://tex.z-dn.net/?f=+%5Csqrt%7Bx%2B1%7D-+%5Csqrt%7Bx-7%7D%3D2)
Ответы
Автор ответа:
0
ОДЗ:
![\left \{ {{x+1 \geq 0} \atop {x-7 \geq 0}} \right. \ =\ \textgreater \ \left \{ {{x \geq -1} \atop {x \geq 7}} \right. \ \ = \ \textgreater \ x \in [7; +\infty) \left \{ {{x+1 \geq 0} \atop {x-7 \geq 0}} \right. \ =\ \textgreater \ \left \{ {{x \geq -1} \atop {x \geq 7}} \right. \ \ = \ \textgreater \ x \in [7; +\infty)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%2B1+%5Cgeq+0%7D+%5Catop+%7Bx-7+%5Cgeq+0%7D%7D+%5Cright.++%5C+%3D%5C+%5Ctextgreater+%5C++%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+-1%7D+%5Catop+%7Bx+%5Cgeq+7%7D%7D+%5Cright.++%5C+%5C+%3D+%5C+%5Ctextgreater+%5C++x+%5Cin+%5B7%3B+%2B%5Cinfty%29)
![( \sqrt{x+1}- \sqrt{x-7})^2=2 ^2 \\ \\ x+1+x-7-2 \sqrt{(x+1)(x-7)}=4 \\ 2 \sqrt{(x+1)(x-7)}=2x-10 \\ 2 \sqrt{(x+1)(x-7)}=2(x-5) \ (:2) \\ \sqrt{(x+1)(x-7)}=x-5 \\ \\ (x+1)(x-7)=x^2-10x+25 \\ x^2-7x+x-7=x^2-10x+25 \\ 4x=32 \\ x=8 ( \sqrt{x+1}- \sqrt{x-7})^2=2 ^2 \\ \\ x+1+x-7-2 \sqrt{(x+1)(x-7)}=4 \\ 2 \sqrt{(x+1)(x-7)}=2x-10 \\ 2 \sqrt{(x+1)(x-7)}=2(x-5) \ (:2) \\ \sqrt{(x+1)(x-7)}=x-5 \\ \\ (x+1)(x-7)=x^2-10x+25 \\ x^2-7x+x-7=x^2-10x+25 \\ 4x=32 \\ x=8](https://tex.z-dn.net/?f=%28+%5Csqrt%7Bx%2B1%7D-+%5Csqrt%7Bx-7%7D%29%5E2%3D2+%5E2+%5C%5C++%5C%5C+x%2B1%2Bx-7-2+%5Csqrt%7B%28x%2B1%29%28x-7%29%7D%3D4+%5C%5C++2+%5Csqrt%7B%28x%2B1%29%28x-7%29%7D%3D2x-10+%5C%5C+2+%5Csqrt%7B%28x%2B1%29%28x-7%29%7D%3D2%28x-5%29+%5C+%28%3A2%29+%5C%5C++%5Csqrt%7B%28x%2B1%29%28x-7%29%7D%3Dx-5+%5C%5C++%5C%5C+%28x%2B1%29%28x-7%29%3Dx%5E2-10x%2B25+%5C%5C+x%5E2-7x%2Bx-7%3Dx%5E2-10x%2B25+%5C%5C+4x%3D32+%5C%5C+x%3D8)
ОТВЕТ: x=8
ОТВЕТ: x=8
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