Решите задачу:
Первый случай a+b=0 решений не дает, так как a и b неотрицательны и одновременно не обращаются в ноль.
Проверка:
Ответ:
![2\sqrt{(x-1)(2-x)}=\sqrt{x-1}+\sqrt{2-x}+(x-1)-(2-x)-[(x-1)+(2-x)];\\\sqrt{x-1}=a\ge 0;\ \sqrt{2-x}=b\ge 0;\ 2ab=a+b+(a^2-b^2)-(a^2+b^2);\\a^2+2ab+b^2=(a+b)+(a-b)(a+b);\ (a+b)^2-(a+b)(1+a-b)=0;\\(a+b)[a+b-(1+a-b)]=0;\ (a+b)(2b-1)=0;\ \left [ {{a+b=0} \atop {2b=1}} \right.](https://tex.z-dn.net/?f=2%5Csqrt%7B%28x-1%29%282-x%29%7D%3D%5Csqrt%7Bx-1%7D%2B%5Csqrt%7B2-x%7D%2B%28x-1%29-%282-x%29-%5B%28x-1%29%2B%282-x%29%5D%3B%5C%5C%5Csqrt%7Bx-1%7D%3Da%5Cge%200%3B%5C%20%5Csqrt%7B2-x%7D%3Db%5Cge%200%3B%5C%202ab%3Da%2Bb%2B%28a%5E2-b%5E2%29-%28a%5E2%2Bb%5E2%29%3B%5C%5Ca%5E2%2B2ab%2Bb%5E2%3D%28a%2Bb%29%2B%28a-b%29%28a%2Bb%29%3B%5C%20%28a%2Bb%29%5E2-%28a%2Bb%29%281%2Ba-b%29%3D0%3B%5C%5C%28a%2Bb%29%5Ba%2Bb-%281%2Ba-b%29%5D%3D0%3B%5C%20%28a%2Bb%29%282b-1%29%3D0%3B%5C%20%5Cleft%20%5B%20%7B%7Ba%2Bb%3D0%7D%20%5Catop%20%7B2b%3D1%7D%7D%20%5Cright. "2\sqrt{(x-1)(2-x)}=\sqrt{x-1}+\sqrt{2-x}+(x-1)-(2-x)-[(x-1)+(2-x)];\\\sqrt{x-1}=a\ge 0;\ \sqrt{2-x}=b\ge 0;\ 2ab=a+b+(a^2-b^2)-(a^2+b^2);\\a^2+2ab+b^2=(a+b)+(a-b)(a+b);\ (a+b)^2-(a+b)(1+a-b)=0;\\(a+b)[a+b-(1+a-b)]=0;\ (a+b)(2b-1)=0;\ \left [ {{a+b=0} \atop {2b=1}} \right.")
