\(\dfrac{2a}{1-\sqrt{a}}\)

b.

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2 tháng 7 2017

a) \(với\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}=\dfrac{2a+2a\sqrt{a}}{1-a}\)

b)\(vớia>b>0\)

\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}=\dfrac{12a\sqrt{a}+6a\sqrt{b}}{4a-b}\)

a: \(\dfrac{5}{3\sqrt{8}}=\dfrac{5\sqrt{2}}{3\cdot4}=\dfrac{5\sqrt{2}}{12}\)

\(\dfrac{2}{\sqrt{b}}=\dfrac{2\sqrt{b}}{b}\)

b: \(\dfrac{5}{5-2\sqrt{3}}=\dfrac{25+10\sqrt{3}}{13}\)

\(\dfrac{2a}{1-\sqrt{a}}=\dfrac{2a\left(1+\sqrt{a}\right)}{1-a}\)

c: \(\dfrac{4}{\sqrt{7}+\sqrt{5}}=\dfrac{4\left(\sqrt{7}-\sqrt{5}\right)}{2}=2\sqrt{7}-2\sqrt{5}\)

\(\dfrac{6a}{2\sqrt{a}-\sqrt{b}}=\dfrac{6a\left(2\sqrt{a}+\sqrt{b}\right)}{4a-b}\)

a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)

b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)

\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)

c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)

15 tháng 9 2018

\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{5-2\sqrt{5}+1}{5-1}=\dfrac{2\left(3-\sqrt{5}\right)}{4}=\dfrac{3-\sqrt{5}}{2}\)

b: \(\dfrac{37}{7+2\sqrt{3}}=7-2\sqrt{3}\)

c:\(=\dfrac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{5}\right)}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\)

d: \(=\dfrac{\left(1+\sqrt{a}\right)\cdot\left(2+\sqrt{a}\right)}{4-a}\)

a: \(=-xy\cdot\dfrac{\sqrt{xy}}{x}=-y\sqrt{yx}\)

b: \(=\sqrt{\dfrac{-105x^3}{35^2}}=\sqrt{-105x}\cdot\dfrac{x}{35}\)

c: \(=\sqrt{\dfrac{5a^3b}{49b^2}}=\sqrt{5ab}\cdot\dfrac{a}{7b}\)

d: \(=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3}\cdot\sqrt{xy}\)

28 tháng 11 2017

a. 2√5

19 tháng 7 2018

câu a nè:

http://123link.pw/0Qyw5v

19 tháng 7 2018

câu d nè : http://123link.pw/Jx46C

nhớ cho đúng nha ^-^

a: \(=2ab\cdot\dfrac{-15}{b^2a}=\dfrac{-30}{b}\)

b: \(=\dfrac{2}{3}\cdot\left(1-a\right)=\dfrac{2}{3}-\dfrac{2}{3}a\)

c: \(=\dfrac{\left|3a-1\right|}{\left|b\right|}=\dfrac{3a-1}{b}\)

d: \(=\left(a-2\right)\cdot\dfrac{a}{-\left(a-2\right)}=-a\)