đưa thừa số vào trong dấu căn
a) \(\frac{x-y}{x}\times\sqrt{\frac{x}{x-y}}\) (x>0: x>y)
b) \(\frac{x+y}{x-y}\times\sqrt{\frac{x-y}{x+y}}\) (x>0: x.y)
c) \(\frac{^{x^2}}{x-5}\times\sqrt{\frac{x-5}{3x}}\) (x>5)
d) \(-2\sqrt{-a}\)
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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
a)\(=-\sqrt{\left(\frac{a}{b}\right)^2\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a^2}{b^2}\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a}{b}}\)
ta có:\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\)
vậy.....
\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right).\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\frac{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)( đpcm )
a, \(\sqrt{\frac{\left(x-y\right)^2}{x^2}\cdot\frac{x}{x-y}}=\) \(\frac{x-y}{x}\)
b. \(\sqrt{\frac{\left(x+y\right)^2}{\left(x-y\right)^2}\cdot\frac{x-y}{x+y}}=\sqrt{\frac{x+y}{x-y}}\)
c.\(\sqrt{\frac{x^4}{\left(x-5\right)^2}\cdot\frac{x-5}{3x}}=\sqrt{\frac{x^3}{3\left(x-5\right)}}\)
\(-2\sqrt{-a}=\sqrt{\left(-2\right)^2\cdot-a}=\sqrt{-4a}\)