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Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{1-x}=b\end{matrix}\right.\) ( \(a,b>0\) )
Ta có: \(\left\{{}\begin{matrix}a^2-b^2=x+1-1+x=2x\\a^2+b^2=x+1+1-x=2\end{matrix}\right.\)
Do đó pt \(\Leftrightarrow\left(a-1\right)\left(b+1\right)=a^2-b^2\)
\(\Leftrightarrow ab+a-b-\frac{a^2+b^2}{2}=a^2-b^2\)
\(\Leftrightarrow2ab+2a-2b-a^2-b^2=2a^2-2b^2\)
\(\Leftrightarrow3a^2-b^2-2ab-2a+2b=0\)
\(\Leftrightarrow3a^2-3ab+ab-b^2-2\left(a-b\right)=0\)
\(\Leftrightarrow3a\left(a-b\right)+b\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(3a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\3a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\3\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-24}{25}\end{matrix}\right.\) ( t/m )
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