1. ĐKXĐ : \(x\ge1\) phương trình tương đương
\(\left(\sqrt{x-1}+1\right)^3+x-1+2\sqrt{x-1}+1=2\)
\(\Leftrightarrow\left(\sqrt{x-1}+1\right)^3+\left(\sqrt{x-1}+1\right)^2=2\)
do \(\sqrt{x-1}\ge0\Rightarrow\sqrt{x-1}+1\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{x-1}+1\right)^3\ge1\\\left(\sqrt{x-1}+1\right)^2\ge1\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x-1}+1\right)^3+\left(\sqrt{x-1}+1\right)^2\ge2\)
Vậy dấu bằng xảy ra khi \(\sqrt{x-1}=0\)\(\Leftrightarrow x=1\)
ĐKXĐ: \(x\le2\)
Phương trình tương đương
\(x^2-1+\sqrt{2-x}-1=2x\sqrt{2-x}-2\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-\dfrac{x-1}{\sqrt{2-x}+1}=\dfrac{2\left(x-1\right)\left(-x^2+x+1\right)}{\sqrt{2-x}+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1-\dfrac{1}{\sqrt{2-x}+1}+\dfrac{x^2-x-1}{\sqrt{2-x}+1}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1+\dfrac{\left(x+1\right)\left(x-2\right)}{\sqrt{2-x}+1}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(1+\dfrac{x-2}{\sqrt{2-x}+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\\dfrac{x-2}{\sqrt{2-x}+1}=-1\end{matrix}\right.\)
\(\dfrac{x-2}{\sqrt{2-x}+1}=-1\Leftrightarrow x-2=-1-\sqrt{2-x}\)
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