tính
C = \(\frac{\sqrt{x}}{\sqrt{x}-1}\)+\(\frac{x+\sqrt{x}}{2}\).( \(\frac{1}{x-\sqrt{x}}\)-\(\frac{1}{x+\sqrt{x}}\))
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24 tháng 7 2017
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
LM
8 tháng 11 2020
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
LM
9 tháng 11 2020
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
\(C=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{x+\sqrt{x}}{2}\left(\frac{x+\sqrt{x}}{\left(x+\sqrt{x}\right)\left(x-\sqrt{x}\right)}-\frac{x-\sqrt{x}}{\left(x+\sqrt{x}\right)\left(x-\sqrt{x}\right)}\right)\)
\(C=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{x+\sqrt{x}}{2}.\frac{2\sqrt{x}}{x^2-x}=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2x\left(\sqrt{x}+1\right)}{2x\left(x-1\right)}=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x-1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}}{x\sqrt{x}-x-\sqrt{x}+1}+\frac{x-1}{x\sqrt{x}-x-\sqrt{x}+1}=\frac{x\sqrt{x}+x-\sqrt{x}-1}{x\sqrt{x}-x-\sqrt{x}+1}\)
\(=\frac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}=\frac{\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)