K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

15 tháng 11 2018

a: \(=5\cdot5\sqrt{3}-\dfrac{1}{3}\cdot3\sqrt{3}=24\sqrt{3}\)

b: \(=\dfrac{12\left(3+\sqrt{5}\right)}{4}=9+3\sqrt{5}\)

c: \(=3-\sqrt{5}+\sqrt{5}=3\)

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

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

a) \(\dfrac{a}{3\sqrt{a}-1}=\dfrac{a\left(3\sqrt{a}+1\right)}{9a-1}\)

15 tháng 11 2023

a: \(\dfrac{\sqrt{5}}{\sqrt{7}}=\dfrac{\sqrt{5\cdot7}}{7}=\dfrac{\sqrt{35}}{7}\)

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

2 tháng 10 2021

Uii em cảm ơn ạ:3

26 tháng 8 2017

bài 1) a) \(xy\sqrt{\dfrac{x}{y}}=x\sqrt{y}\sqrt{y}\dfrac{\sqrt{x}}{\sqrt{y}}=x\sqrt{x}\sqrt{y}=\left(\sqrt{x}\right)^3\sqrt{y}\)

b) \(\sqrt{\dfrac{5a^3}{49b}}=\dfrac{\sqrt{5a^3}}{\sqrt{49b}}=\dfrac{\sqrt{5a^3}}{7\sqrt{b}}=\dfrac{\sqrt{5a^3}.\sqrt{b}}{7\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{5a^3b}}{7b}\)

bài 2) a) \(\dfrac{\sqrt{3}-3}{1-\sqrt{3}}=\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=\sqrt{3}\)

b) \(\dfrac{5-\sqrt{15}}{\sqrt{3}-\sqrt{5}}=\dfrac{-\sqrt{5}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}-\sqrt{5}}=-\sqrt{5}\)

c) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)

22 tháng 1 2017

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

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

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: \(\dfrac{6}{5\sqrt{8}}=\dfrac{6}{10\sqrt{2}}=\dfrac{3}{5\sqrt{2}}=\dfrac{3\sqrt{2}}{10}\)

b: \(\dfrac{7}{5+2\sqrt{3}}=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

c: \(\dfrac{6}{\sqrt{7}-\sqrt{5}}=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}=3\left(\sqrt{7}+\sqrt{5}\right)\)

24 tháng 8 2023

a) \(\dfrac{6}{5\sqrt{8}}\)

\(=\dfrac{6}{5\cdot2\sqrt{2}}\)

\(=\dfrac{6}{10\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{10}\)

b) \(\dfrac{7}{5+2\sqrt{3}}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}\)

\(=\dfrac{7\left(5-2\sqrt{3}\right)}{13}\)

\(=\dfrac{35-14\sqrt{3}}{13}\)

c) \(\dfrac{6}{\sqrt{7}-\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)

\(=\dfrac{6\left(\sqrt{7}+\sqrt{5}\right)}{2}\)

\(=3\sqrt{7}+3\sqrt{5}\)

`(a+4 sqrt a +4)/(sqrt a+2)`

`= (sqrt a+2)^2/(sqrt a+2)`

`= sqrt a +2`

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