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30 tháng 6 2019

\(M=\sqrt{\frac{m}{1-2x+x^2}}\times\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m\times\left(1-2x+x^2\right)}}{\sqrt{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m}\times\sqrt{1-2x+x^2}}{9}\)

\(=\frac{\sqrt{m}\times\sqrt{4m}}{9}\)

\(=\frac{2m}{9}\)

vậy . . .

18 tháng 11 2020

\(M=\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

     \(=\sqrt{\frac{m}{\left(1-x\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

     \(=\sqrt{\frac{m}{\left(1-x\right)^2}.\frac{4m\left(1-x\right)^2}{81}}\)

     \(=\frac{\sqrt{4m^2}}{81}\)

     \(=\frac{\sqrt{4m^2}}{\sqrt{81}}=\frac{2m}{9}\)

Vậy : \(M=\frac{2m}{9}\) 

2 tháng 10 2017

\(\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(x-1\right)^2}{81}}=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{2\sqrt{m}.\left|x-1\right|}{9}=\frac{2m}{9}\)

\(x\ne1\) chứ không phải x>1 nên không thể ghi |x-1|=x-1 nhé Despacito

2 tháng 10 2017

A..mk vua nghi ra bai nay

\(\sqrt{\frac{m}{x^2-2x+1}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\sqrt{\frac{m}{\left(x-1\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

\(=\frac{\sqrt{m}}{\left|x-1\right|}.\frac{\sqrt{4m\left(x-1\right)^2}}{9}\) ( Thoa man DKXD \(m>0;x\ne1\)

\(=\frac{\sqrt{m}}{x-1}.\frac{2\left(x-1\right).\sqrt{m}}{9}\)

\(=\frac{2m}{9}\)

ko biet co dung ko nua 

\(\sqrt{\dfrac{m}{1-2x+x^2}}+\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{\left(1-x\right)^2}}+\sqrt{\dfrac{4m\left(1-2x+x^2\right)}{81}}\\ =\dfrac{\sqrt{m}}{\left|1-x\right|}+\dfrac{2\sqrt{m}\left|1-x\right|}{9}=\dfrac{9\sqrt{m}+2\sqrt{m}\left(1-x\right)^2}{\left|1-x\right|.9}\\ =\dfrac{\sqrt{m}\left(9+2-4x+2x^2\right)}{\left(x-1\right).9}\)

tới đây thì hết bt rồi

6 tháng 7 2018

P=\(\frac{2\sqrt{x}}{\sqrt{x}+m}+\frac{2\sqrt{x}}{\sqrt{x}-m}-\)\(\frac{m^2}{4x-4m^2}\)

  = \(\frac{8\sqrt{x}\left(\sqrt{x}-m\right)+8\sqrt{x}\left(\sqrt{x}+m\right)-m^2}{4x-4m^2}\)

=\(\frac{8\sqrt{x}\left(\sqrt{x}-m+\sqrt{x}+m\right)-m^2}{4x-4m^2}\)

=\(\frac{8\sqrt{x}.2\sqrt{x}-m^2}{4x-4m^2}\)

=\(\frac{16x-m^2}{4x-4m^2}\)

3 tháng 6 2019

ĐK:x>1

M=\(\frac{x-1}{2x}\) .\(\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{x-1}\)

=\(\frac{x-1}{2x}\).\(\frac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{x-1}\)=\(\frac{x-1}{2x}\).\(\frac{-4x}{x-1}\)=-2

Vậy M=-2

22 tháng 4 2017

a) \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{a}{b}}\) với a>0 và b>0

b) \(\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m-8mx+4mx^2}{81}}=\sqrt{\dfrac{m}{1-2x+x^2}}.\sqrt{\dfrac{4m\left(2-2x+x^2\right)}{81}}\)

\(=\sqrt{\dfrac{4m^2\left(1-2x+x^2\right)}{81\left(1-2x+x^2\right)}}=\sqrt{\dfrac{4m^2}{81}}=\sqrt{\dfrac{2m}{9}}\)

14 tháng 7 2018

a) \(M=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1\)\(-\frac{2x+\sqrt{x}}{\sqrt{x}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{x-\sqrt{x}+1}\)\(+\frac{\sqrt{x}-2x-\sqrt{x}}{\sqrt{x}}\)

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

\(=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)

28 tháng 7 2019

\(M=\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\left(x< 0;x\ge2\right)\)

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

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

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

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

\(=\frac{4x\sqrt{x^2-2x}}{-2x}=-2x\sqrt{x^2-2x}\)