Question

[MINH GIANG]

Giải phương trình

\(\sqrt{x}+\sqrt{2-x}+\sqrt{2x-x^2}=3\)

Answer 1

[Toru]

\(\sqrt{x}+\sqrt{2-x}+\sqrt{2x-x^2}=3\) (ĐKXĐ: \(0\le x\le2\))

\(\Leftrightarrow\sqrt{x}+\sqrt{2-x}+\sqrt{x\left(2-x\right)}=3\) (1)

Đặt \(\sqrt{x}+\sqrt{2-x}=a\Rightarrow\dfrac{a^2-2}{2}=\sqrt{x\left(2-x\right)}\) (2) (a > 0)

Thay (2) vào (1), ta được:

\(a+\dfrac{a^2-2}{2}=3\)

\(\Leftrightarrow a^2+2a-2=6\)

\(\Leftrightarrow a^2+2a-8=0\) \(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-4\end{matrix}\right.\)

Mà a > 0 nên \(a=2\)

\(\Rightarrow\sqrt{x}+\sqrt{2-x}=2\)

\(\Leftrightarrow x+2-x+2\sqrt{x\left(2-x\right)}=2\)

\(\Leftrightarrow2\sqrt{x\left(2-x\right)}=0\)

\(\Leftrightarrow x\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmdk\right)\\x=2\left(tmdk\right)\end{matrix}\right.\)

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