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19 tháng 7 2020

a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)

b)\(\sqrt{125a^2}=\sqrt{5^2.5.a^2}=5.\left|a\right|\sqrt{5}=-5a\sqrt{5}\) ( vì a<0)

c)\(-\sqrt{500.162}=-\sqrt{10^2.5.9^2.2}=-10.9\sqrt{5.2}=-90\sqrt{10}\)

d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{1}{3}.15.\left|a\right|=\frac{15a}{3}\) ( a>0)

a) Ta có: \(\sqrt{96}\cdot\sqrt{125}\)

\(=\sqrt{16}\cdot\sqrt{6}\cdot\sqrt{25}\cdot\sqrt{5}\)

\(=20\cdot\sqrt{30}\)

b) Ta có: \(\sqrt{a^4\cdot6^5}\)

\(=a^2\cdot36\cdot\sqrt{6}\)

c) Ta có: \(\sqrt{a^6\cdot b^{11}}\)

\(=\sqrt{a^6}\cdot\sqrt{b^{11}}\)

\(=\left|a^3\right|\cdot\left|b^5\right|\cdot\sqrt{b}\)

\(=a^3b^5\cdot\sqrt{b}\)

d) Ta có: \(\sqrt{a^3\left(1-a\right)^4}\)

\(=\sqrt{a^3}\cdot\sqrt{\left(1-a\right)^4}\)

\(=a\sqrt{a}\cdot\left(1-a\right)^2\)

13 tháng 7 2016

a/ \(0,1\sqrt{2.10000=0,1\sqrt{ }2.100^{ }2=0,1\cdot100\sqrt{ }2=10\sqrt{ }2}\) 

b/ \(-0,05\sqrt{28800}=-0,05\sqrt{288\cdot100=-0,05\cdot10\sqrt{ }288=6\sqrt{ }2}\) 

c/\(\sqrt{7\cdot63}a^2=\sqrt{7\cdot9\cdot7}a^2=21a^2\) 

 

13 tháng 7 2016

\(\sqrt{72a^{ }2b\sqrt{ }4=\sqrt{ }6\cdot9\left|\right|ab^{ }2=-3\sqrt{ }6ab^{ }2}\)

19 tháng 10 2021

\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)

19 tháng 10 2021

\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)

\(\sqrt{225\cdot17}=15\sqrt{17}\)

19 tháng 1 2022

a, \(-\frac{2}{3}\sqrt{ab}=-\sqrt{\frac{4ab}{9}}\)

b, \(a\sqrt{\frac{3}{a}}=\sqrt{\frac{3a^2}{a}}=\sqrt{3a}\)

c, \(a\sqrt{7}=\sqrt{7a^2}\)

d, \(b\sqrt{3}=\sqrt{3b^2}\)

e, \(ab\sqrt{\frac{a}{b}}=\sqrt{\frac{a^3b^2}{b}}=\sqrt{a^3b}\)

f, \(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a^2b^2}{a}+\frac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)

3 tháng 10 2022

a, −23ab=−4ab9

b, a3a=3a2a=3a

c, a7=7a2

d, b3=3b2

e, abab=a3b2b=a3b

f, 

 

18 tháng 1 2022

chịu.-.

HT~~~

3 tháng 10 2022

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13 tháng 7 2016

a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)

\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)

\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)

\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)

b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)

\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)

C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)

\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)

\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)

\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)

d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)

\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)

e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)

\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)

\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)

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

 

7 tháng 10 2019

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

15 tháng 8 2021

đưa thừa số vào trong dấu căn*

mình nhầm

a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)

b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)

\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)