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Ta có : A = \(\left(\frac{x+2}{x.\sqrt{x}-1}+\frac{\sqrt{x}+2}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
= \(\frac{x+2+x+\sqrt{x}-2-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
= \(\frac{x-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=1\)
Vậy A = 1
\(A=\frac{\sqrt{x}+1}{x-1}-\frac{x+2}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)\(-\frac{x-2}{\sqrt{x}^3-1}\)\(-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}-1}\)\(-\frac{x-2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)\(-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x+\sqrt{x}+1-x+2-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{2}{\sqrt{x}^3-1}\)
ĐK : x>0, x khác 1
\(A=\left(\frac{1}{\sqrt{x}+1}+\frac{2\left(1-\sqrt{x}\right)}{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{2}{x-1}\right)\)
\(=\left(\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\frac{1}{\sqrt{x}+\sqrt{x-1}}-\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}-\frac{\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}-\frac{x\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
Bạn khử mẫu , xong rồi quy đồng là xong