Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\frac{2xy}{2\sqrt{x}+3\sqrt{y}}=\frac{2xy.\left(2\sqrt{x}-3\sqrt{y}\right)}{\left(2\sqrt{x}+3\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{4x\sqrt{x}y-6xy\sqrt{y}}{2x-3y}\)
\(b,\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\sqrt{x}}{2\sqrt{x}\sqrt{x}}=\frac{x+\sqrt{xy}}{2x}\)
\(c,\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{2.\left(\sqrt{3}-1\right)}{2}=\sqrt{3}-1\)
\(d,\frac{6}{2\sqrt{3}+\sqrt{2}}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{10}=\frac{6\sqrt{3}-3\sqrt{2}}{5}\)
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
a/ \(\frac{\sqrt{a}-\left(\sqrt{a}\right)^2}{\sqrt{a}-1}\)
=\(\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{\sqrt{a}-1}\)
=\(\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\)
=\(-\sqrt{a}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
a/ \(\frac{\sqrt{2}-\sqrt{3}}{2\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{2}-\sqrt{3}\right)}{2.3}=\frac{\sqrt{6}-\sqrt{9}}{6}=\frac{\sqrt{6}-3}{6}\)
b/ \(\frac{x+a\sqrt{x}}{a\sqrt{x}}=\frac{\sqrt{x}\left(x+a\sqrt{x}\right)}{a.\left|x\right|}=\frac{x\sqrt{x}+a\left|x\right|}{a\left|x\right|}\)
c/ \(\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{x}-\sqrt{y}\right)}{\left|x\right|-\left|y\right|}\)
d/ \(\frac{x}{2\sqrt{x}-3\sqrt{y}}=\frac{x\left(2\sqrt{x}+3\sqrt{y}\right)}{\left(2\sqrt{x}-3\sqrt{y}\right)\left(2\sqrt{x}+3\sqrt{y}\right)}=\frac{2x\sqrt{x}+3x\sqrt{y}}{4\left|x\right|-9\left|y\right|}\)