\(\sqrt{9\left(x-5\right)^2}\) (\(x\ge5\))...">
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8 tháng 7 2019

a) \(\sqrt{9\left(x-5\right)^2}=9\left|x-5\right|=9\left(x-5\right)=9x-45\)

b) \(\sqrt{x^2\left(x-2\right)^2}=\left|x\left(x-2\right)\right|\)

Nhận xét rằng x < 0 thì x - 2 bé hơn 0 suy ra x (x -2 ) > 0

Suy ra \(\sqrt{x^2\left(x-2\right)^2}=\left|x\left(x-2\right)\right|=x\left(x-2\right)\)
Làm gấp chẳng biết đúng hay sai.

8 tháng 7 2019

\(\sqrt{9\left(x-5\right)^2}=\sqrt{3^2\left(x-5\right)^2}=3\left|x-5\right|\)tth sai r

19 tháng 9 2018

\(\sqrt{72a^8\left(x^2-4x+4\right)}=\sqrt{72a^8\left(x-2\right)^2}=\sqrt{72}a^4|\left(x-2\right)|=\sqrt{72}a^4\left(2-x\right)\)

19 tháng 9 2018

\(\sqrt{40x^6\left(a^2+6a+9\right)}=\sqrt{40x^6\left(x+3\right)^2}=\sqrt{40}|x^3\left(x+3\right)|=\sqrt{40}.\left(-x^3\right)\left(3-x\right)\)

\(=-\sqrt{40}x^3\left(3-x\right)\)

18 tháng 8 2017

giúp mình vs nha...please!!!..

18 tháng 8 2017

a. 1/2*(-x^5)

b. (10-x)^5

c. x-4+(4-x) = 0

d. 6-2x-(3-x) = 3-x

17 tháng 8 2018

\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)

\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)

\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)

\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)

\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)

\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)

26 tháng 8 2020

a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)

\(=x+3+\sqrt{\left(x-3\right)^2}\)

\(=x+3+\left|x-3\right|\)

\(=x+3-\left(x-3\right)\)

\(=x+3-x+3\)

\(=6\)

b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)

\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)

\(=\left|x+2\right|-\left|x\right|\)

\(=x+2-\left(-x\right)\)

\(=x+2+x\)

\(=2x+2=2\left(x+1\right)\)

c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)

\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)

\(=\frac{\left|x-1\right|}{x-1}\)

\(=\frac{x-1}{x-1}=1\)

d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)

\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)

\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)

\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)

\(=\left|x-2\right|-1\)

\(=-\left(x-2\right)-1\)

\(=-x+2-1\)

\(=-x+1=-\left(x-1\right)\)

19 tháng 10 2018

1/ Thực hiện phép tính

a) 9220+12235

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

\(=\sqrt{5}-\sqrt{4}+\sqrt{7}-\sqrt{5}=\sqrt{7}-\sqrt{4}=\sqrt{7}-2\)