A= (\(\frac{1}{x-\sqrt{x}}\)- \(\frac{\sqrt{x}}{\sqrt{x}-1}\)) :(\(\frac{2}{x-1}\) - \(\frac{1}{\sqrt{x}+1}\))
a) rút gọn biểu thức A
b) chứng minh A-2>0 với mọi x thỏa mãn đkiện x>0 và x # 0
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10 tháng 2 2020
a, Ta có : \(A=\left(\frac{x-\sqrt{x}+2}{x-1}-\frac{1}{\sqrt{x}-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-\sqrt{x}+2-\left(\sqrt{x}+1\right)}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-2\sqrt{x}+1}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{\left(\sqrt{x}-1\right)^2}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}\frac{\left(x+2\sqrt{x}\right)}{\left(2x-2\sqrt{x}\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(2x-2\sqrt{x}\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)2\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{\sqrt{x}+2}{2\sqrt{x}+2}\)
b, Ta có : \(A=\frac{\sqrt{x}+1+1}{2\left(\sqrt{x}+1\right)}=\frac{1}{2}+\frac{1}{2\left(\sqrt{x}+1\right)}\)
- Ta thấy : \(\sqrt{x}+1>0\)
=> \(\frac{1}{2\left(\sqrt{x}+1\right)}>0\)
=> \(\frac{1}{2\left(\sqrt{x}+1\right)}+\frac{1}{2}>\frac{1}{2}\)
=> \(A>\frac{1}{2}\) ( đpcm )
2 tháng 7 2020
Bài 1:
\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
13 tháng 7 2020
a) Ta có: \(P=\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}}{x+1}\right)\)
\(=\left(\frac{x+1}{\left(\sqrt{x}-1\right)\left(x+1\right)}-\frac{2}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right):\left(\frac{x+1}{x+1}-\frac{\sqrt{x}}{x+1}\right)\)
\(=\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}:\frac{x+1-\sqrt{x}}{x+1}\)
\(=\frac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\frac{x+1}{x-\sqrt{x}+1}\)
\(=\frac{x-1}{\sqrt{x}-1}\cdot\frac{1}{x-\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+1}{x-\sqrt{x}+1}\)