Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{6}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\sqrt{6}^2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}}{2\sqrt{6}-1}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{6}+1\right)}{2\sqrt{6}^2-1^2}=\frac{4\sqrt{3}+6\sqrt{2}+12+\sqrt{2}+\sqrt{3}+\sqrt{6}}{11}\)\(=\frac{\sqrt{6}+5\sqrt{3}+7\sqrt{2}+12}{11}\)
\(b,\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{z}+\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)}=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{\left(\sqrt{x}+\sqrt{y}\right)^2-\sqrt{z}^2}\)
\(=\frac{\sqrt{x}+\sqrt{y}-\sqrt{z}}{x+2\sqrt{xy}+y-z}\)
\(a,\frac{2xy}{2\sqrt{x}+3\sqrt{y}}=\frac{2xy.\left(2\sqrt{x}-3\sqrt{y}\right)}{\left(2\sqrt{x}+3\sqrt{y}\right)\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{4x\sqrt{x}y-6xy\sqrt{y}}{2x-3y}\)
\(b,\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\sqrt{x}}{2\sqrt{x}\sqrt{x}}=\frac{x+\sqrt{xy}}{2x}\)
\(c,\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{2.\left(\sqrt{3}-1\right)}{2}=\sqrt{3}-1\)
\(d,\frac{6}{2\sqrt{3}+\sqrt{2}}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}=\frac{6\left(2\sqrt{3}-\sqrt{2}\right)}{10}=\frac{6\sqrt{3}-3\sqrt{2}}{5}\)
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/ \(\frac{x}{2\sqrt{x}-3\sqrt{y}}=\frac{x\left(2\sqrt{x}+3\sqrt{y}\right)}{\left(2\sqrt{x}-3\sqrt{y}\right)\left(2\sqrt{x}+3\sqrt{y}\right)}=\frac{2x\sqrt{x}+3x\sqrt{y}}{4\left|x\right|-9\left|y\right|}\)
\(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2\left(\sqrt[3]{4}-\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\left(\sqrt[3]{4^2}+\sqrt[3]{4}.\sqrt[3]{2}+\sqrt[3]{2^2}\right)}\)
\(=\frac{2\left(\sqrt[3]{4}-\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}\right)^3-\left(\sqrt[3]{2}\right)^3}=\sqrt[3]{4}-\sqrt[3]{2}\)
\(y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{2\left(\sqrt[3]{4}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}+\sqrt[3]{2}\right)\left(\sqrt[3]{4^2}-\sqrt[3]{4}.\sqrt[3]{2}+\sqrt[3]{2^2}\right)}\)
\(=\frac{6\left(\sqrt[3]{4}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}\right)^3+\left(\sqrt[3]{2}\right)^3}=\sqrt[3]{4}+\sqrt[3]{2}\)
\(P=\frac{xy}{x+y}=\frac{\sqrt[3]{4^2}-\sqrt[3]{2^2}}{2\sqrt[3]{4}}=\frac{\sqrt[3]{4}-1}{2}\)
\(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
\(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}+\sqrt{7}\right)\left(\sqrt{10}-\sqrt{7}\right)}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\dfrac{3\left(\sqrt{10}-\sqrt{7}\right)}{3}=\sqrt{10}-\sqrt{7}\)
\(\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\dfrac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-1-2}=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}=\sqrt{x-1}+\sqrt{2}\)