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\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)Đkxđ : x>2
=(\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) \(:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)-\left(x-4\right)}\)
\(=\frac{1}{\sqrt{x}}.\frac{\sqrt{x}-2}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(\left(x-1\right)-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(=\left(x-1\right)-\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=x-1-2\sqrt{x}+\sqrt{x}+1\)
\(=x-\sqrt{x}\)
Đk: x > 0, x khác 1
Làm ngắn gọn thôi nhé, bạn tự khai triển ra
\(A=\frac{x+2}{x\sqrt{x}-1}+\left(\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right)=\frac{x+2}{x\sqrt{x}-1}+\frac{-\sqrt{x}-2}{\sqrt{x}^3-1}\)
\(\frac{\left(\sqrt{x}-1\right)x^2-x+\sqrt{x}}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}^3-1\right)}\)(Chú ý \(x\sqrt{x^3}=x^2\sqrt{x},\sqrt{x^3}=\left|x\right|\sqrt{x}=x\sqrt{x}\left(x>0\right)\)
Tử = \(\sqrt{x}\left[\left(\sqrt{x}-1\right)x\sqrt{x}-\left(\sqrt{x}-1\right)\right]=\sqrt{x}\left(\sqrt{x}-1\right)\left(x\sqrt{x}-1\right)\)
Mẫu = ....
Rồi giản ước. Kết quả là \(A=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)