Question

đặt ẩn phụ 

Answer 1

\(ĐK:0\le x\le7\)

Đặt \(\left\{{}\begin{matrix}\sqrt{7-x}=a\ge0\\\sqrt{x}=b\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=7\\ab=\sqrt{7x-x^2}\end{matrix}\right.\), PTTT:

\(b^2+2a+ab-2b-a^2-b^2=0\\ \Leftrightarrow a\left(b-a\right)-2\left(b-a\right)=0\\ \Leftrightarrow\left(b-a\right)\left(a-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\a=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7-x\\7-x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\)