a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\left(x>0;x\ne1\right)\)
\(=\dfrac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)
\(=\sqrt{x}-1\)
\(---\)
b) Thay \(x=3+2\sqrt{2}\) vào \(A\), ta được:
\(A=\sqrt{3+2\sqrt{2}}-1\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)
\(=\left|\sqrt{2}+1\right|-1\)
\(=\sqrt{2}+1-1\)
\(=\sqrt{2}\)
\(Toru\)
\(a,A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\left(dk:x>0,x\ne1\right)\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
\(b,x=3+2\sqrt{2}=\sqrt{2}^2+2\sqrt{2}.1+1=\left(\sqrt{2}+1\right)^2\)
\(A=\sqrt{x}-1=\sqrt{\left(\sqrt{2}+1\right)}^2-1=\sqrt{2}+1-1=\sqrt{2}\)
