ĐKXĐ: \(x\ge1\)
\(x^2+x-2+4\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)+4\sqrt{x-1}=0\)
Do \(x\ge1\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x+2\right)\ge0\\\sqrt{x-1}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)+4\sqrt{x-1}\ge0\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\sqrt{x-1}=0\end{matrix}\right.\) \(\Rightarrow x=1\)
Cakpan làm thêm giúp mình theo này đc không ạ?
\(\Leftrightarrow x^2=2-x-2\sqrt{x-1}.2\)
\(\Leftrightarrow x^2+x-1+4=2-x+x+1-2\sqrt{x-1}.2+4\)
\(\Leftrightarrow x^2+x+x-1-2+4=\left(\sqrt{x-1}+2\right)^2\)
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