Question

[Ngô quang minh]

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

giải pt

Answer 1

[alibaba nguyễn]

Điều kiện \(\hept{\begin{cases}2+x\ge0\\2-x\ge0\end{cases}}\Leftrightarrow-2\le x\le2\)

Đặt \(\hept{\begin{cases}\sqrt{2+x}=a\left(a\ge0\right)\\\sqrt{2-x}=b\left(b\ge0\right)\end{cases}\Rightarrow a^2+b^2=4}\)thì

\(1PT\Leftrightarrow\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}a^2+\sqrt{2}b^2-a^2b+ab^2=2\sqrt{2}-2b+2a-\sqrt{2}ab\)

\(\Leftrightarrow2\sqrt{2}-a^2b+ab^2+2b-2a+\sqrt{2}ab=0\)

\(\Leftrightarrow\sqrt{2}\left(2+ab\right)+ab\left(b-a\right)+2\left(b-a\right)=0\)

\(\Leftrightarrow\sqrt{2}\left(2+ab\right)+\left(b-a\right)\left(2+ab\right)=0\)

\(\Leftrightarrow\left(2+ab\right)\left(\sqrt{2}+b-a\right)=0\)

\(\Leftrightarrow a-b=\sqrt{2}\)(vì 2 + ab > 0)

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

\(\Leftrightarrow4-2\sqrt{4-x^2}=2\)

\(\Leftrightarrow\sqrt{4-x^2}=1\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\left(l\right)\end{cases}}\)

Answer 2

[Ngô quang minh]

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