2.trục căn thức ở mẫu sau:
a\(\frac{2xy}{2\sqrt{x}+3\sqrt{y}}\)
b\(\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}}\)
c\(\frac{2}{\sqrt{3}+1}\)
d\(\frac{6}{2\sqrt{3}+\sqrt{2}}\)
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NQ
5 tháng 8 2019
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
B
23 tháng 7 2019
\(a,\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\sqrt{6}^2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{2\sqrt{6}^2-1^2}=\frac{4\sqrt{3}+6\sqrt{2}+12+\sqrt{2}+\sqrt{3}+\sqrt{6}}{11}\)\(=\frac{\sqrt{6}+5\sqrt{3}+7\sqrt{2}+12}{11}\)
\(b,\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{z}+\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{z}^2}\)
\(=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{x+2\sqrt{xy}+y-z}\)
NV
18 tháng 7 2016
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/ \(\frac{x}{2\sqrt{x}-3\sqrt{y}}=\frac{x\left(2\sqrt{x}+3\sqrt{y}\right)}{\left(2\sqrt{x}-3\sqrt{y}\right)\left(2\sqrt{x}+3\sqrt{y}\right)}=\frac{2x\sqrt{x}+3x\sqrt{y}}{4\left|x\right|-9\left|y\right|}\)
\(a,\frac{2xy}{2\sqrt{x}+3\sqrt{y}}=\frac{2xy.\left(2\sqrt{x}-3\sqrt{y}\right)}{\left(2\sqrt{x}+3\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{4x\sqrt{x}y-6xy\sqrt{y}}{2x-3y}\)
\(b,\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\sqrt{x}}{2\sqrt{x}\sqrt{x}}=\frac{x+\sqrt{xy}}{2x}\)
\(c,\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{2.\left(\sqrt{3}-1\right)}{2}=\sqrt{3}-1\)
\(d,\frac{6}{2\sqrt{3}+\sqrt{2}}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{10}=\frac{6\sqrt{3}-3\sqrt{2}}{5}\)