\(\frac{1+\frac{4}{x-2}}{\frac{x^2-4}{2}}\)

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28 tháng 3 2020

=\(\left(1+\frac{4}{x-2}\right):\left(\frac{x^2-4}{2}\right)\)

=\(\left(\frac{x-2}{x-2}+\frac{4}{x-2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{2}\right)\)

=\(\frac{x-2+4}{x-2}\cdot\frac{2}{\left(x-2\right)\left(x+2\right)}\)

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

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

3 tháng 4 2020

Mình quên ĐKXĐ 

Sorry bạn nha

\(\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)

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

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

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

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

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

\(=\frac{x-2}{1}\times\frac{x+2}{2}=\frac{\left(x-2\right)\left(x+2\right)}{1.2}=\frac{x^2-2^2}{2}=\frac{x^2-2}{1}=x^2-2\)

(Sai thì thôi)

#Học tốt!!!

~NTTH~

24 tháng 12 2019

Mơn bạn nha☺☺☺

7 tháng 8 2016

ĐKXĐ: \(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)

\(P=\left(\frac{x+1}{x-2}-\frac{2x}{x+2}-\frac{x^2-x}{4-x^2}\right):\left(3-\frac{3x+4}{x+2}\right)\)

\(=\left[\frac{x+1}{x-2}-\frac{2x}{x+2}+\frac{x^2-x}{\left(x+2\right)\left(x-2\right)}\right]:\left(3-\frac{3x+4}{x+2}\right)\)

\(=\left[\frac{\left(x+1\right)\left(x+2\right)-2x\left(x-2\right)+x^2-x}{\left(x+2\right)\left(x-2\right)}\right]:\left(\frac{3x+6-3x-4}{x+2}\right)\)

\(=\left(\frac{x^2+3x+2-2x^2+4x+x^2-x}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{2}\)

\(=\frac{6x+2}{x-2}.\frac{1}{2}=\frac{3x+1}{x-2}\)

16 tháng 12 2016

Đào Thị Phương Duyên mk trả lời câu kia của pn rùi ak

16 tháng 12 2016

nhẩm cx ra (x+1)2 / x2 -4

11 tháng 12 2018

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

11 tháng 12 2018

Bạn có thể giúp mình 2 câu còn lại dc kh ạ 

3 tháng 8 2016

Đề phần a sai

3 tháng 8 2016

bạn sử hộ mình

 

11 tháng 12 2020

a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))

\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)

\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)

\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)

\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)

\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)

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

\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)

b) Với \(M=\frac{1}{4}\)ta có :

\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)

\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)

Vậy x = 3

11 tháng 12 2020

a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)

\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)

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

b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)

30 tháng 11 2016

a. 2x

b.\({3x}\over x^2-1\)