ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}+1-\sqrt{x^3+x^2+x+1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\) (1)
Do \(x\ge1\Rightarrow\sqrt{x^3+x^2+x+1}\ge\sqrt{1+1+1+1}=2\)
\(\Rightarrow\sqrt{x^3+x^2+x+1}-1>0\)
Do đó (1) tương đương:
\(\sqrt{x-1}-1=0\Rightarrow x=2\)
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