Question

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Answer 1

\(A=\sqrt{\left(\sqrt{2}+1\right)^2}-\dfrac{\sqrt{2}^2}{\sqrt{2}}=\left|\sqrt{2}+1\right|-\sqrt{2}=\sqrt{2}+1-\sqrt{2}=1\)

\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=x-1\)

b.

\(B+\sqrt{2}\ge\sqrt{2}x\Leftrightarrow x-1+\sqrt{2}\ge\sqrt{2}x\)

\(\Leftrightarrow\left(\sqrt{2}-1\right)x\le\sqrt{2}-1\)

\(\Rightarrow x\le1\)

Kết hợp ĐKXĐ ta được: \(0\le x< 1\)