ĐKXĐ: \(x\ge2\sqrt{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-8}=a\ge0\\\sqrt{x-2}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+1=ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2=2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)
\(\Leftrightarrow a=b=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x^2-8}=1\\\sqrt{x-2}=1\end{matrix}\right.\) \(\Leftrightarrow x=3\)
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