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

\(=\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+2\right)^2-\left(x-2\right)^2-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+4x+4-x^2+4x-4-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8x-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8}{x+2}\)

bạn ơi, cho mình hỏi

4-x² thì sẽ = 2²-x² = (2-x)(2+x)

vậy sao bạn ghi là (x-2)(x+2) vậy???

x=50 thư chị hà

bài dễ ợt,mỗi vế trừ đi 2

bạn ơi đề thiếu

Theo đề ra ,ta có :

    - 1 / 12 < x < 1 / 8 mà x có giá trị nguyên 

=> x = 0

1.Vì  \(\frac{x}{-2}=\frac{-8}{x}\Rightarrow-2.\left(-8\right)=x.x\)

                                        \(16=x.x\)hay \(4^2=x^2\Rightarrow x=4\)

2. Rút gọn : \(\frac{20}{28}=\frac{5}{7}=\frac{-5}{-7}\)

\(\Rightarrow x=-7\)

3. \(\frac{x}{2}-\frac{11}{5}=\frac{7}{8}\times\frac{64}{49}\)

\(\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)

Mà \(\frac{8}{7}+\frac{11}{5}=\frac{502}{35}\)

\(\Rightarrow x=\frac{234}{35}\)

1) \(\frac{x}{-2}=\frac{-8}{x}\)

\(\Rightarrow x\times x=\left(-2\right)\times\left(-8\right)\)

\(\Rightarrow x^2=16\)

\(\Rightarrow\orbr{\begin{cases}x^2=4^2\\x^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

Vậy x = 4 hoặc x = -4

2) \(\frac{-5}{x}=\frac{20}{28}\)

\(\Rightarrow\frac{-5}{x}=\frac{5}{7}\)

\(\Rightarrow5\times x=\left(-5\right)\times7\)

\(\Rightarrow5\times x=-35\)

\(\Rightarrow x=\left(-35\right):5\)

\(\Rightarrow x=-7\)

Vậy x = -7

3) \(\frac{x}{2}-\frac{11}{5}=\frac{7}{8}\times\frac{64}{49}\)

\(\Rightarrow\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)

\(\Rightarrow\frac{x}{2}=\frac{8}{7}+\frac{11}{5}\)

\(\Rightarrow\frac{x}{2}=\frac{117}{35}\)

\(\Rightarrow35x=117\times2\)

\(\Rightarrow35x=234\)

\(\Rightarrow x=234:35\)

\(\Rightarrow x=\frac{234}{35}\)

Vậy  \(x=\frac{234}{35}\)

4) \(\frac{x}{5}+\frac{9}{2}=\frac{6}{7}\times\frac{36}{48}\)

\(\Rightarrow\frac{x}{5}+\frac{9}{2}=\frac{9}{14}\)

\(\Rightarrow\frac{x}{5}=\frac{9}{14}-\frac{9}{2}\)

\(\Rightarrow\frac{x}{5}=\frac{-27}{7}\)

\(\Rightarrow7x=\left(-27\right)\times5\)

\(\Rightarrow7x=-135\)

\(\Rightarrow x=\left(-135\right):7\)

\(\Rightarrow x=\frac{-135}{7}\)

Vậy  \(x=\frac{-135}{7}\)

5) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow\frac{3}{x-5}+\frac{4}{x+2}=0\)

\(\Rightarrow3\left(x+2\right)+4\left(x-5\right)=0\)

\(\Rightarrow3x+6+4x-20=0\)

\(\Rightarrow\left(3x+4x\right)+\left(6-20\right)=0\)

\(\Rightarrow7x-14=0\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=14:7\)

\(\Rightarrow x=2\)

Vậy x = 2

_Chúc bạn học tốt_

\(\frac{1}{\left(x+1\right)^2\left(x+2\right)}=\frac{a}{x+1}+\frac{b}{\left(x+1\right)^2}+\frac{c}{x+2}\)

\(=\frac{a}{x+1}+\frac{b}{x+1^2}+\frac{c}{x+2}\)

\(=\frac{1}{\left(x+1\right)^2\left(x+2\right)=}=\frac{a}{\left(x+1\right)\left(x+2\right)}+\frac{b}{x+2}+\frac{c}{\left(x+1\right)^2\left(x+2\right)}\)

\(\frac{c}{\left(x+1\right)^2}+\frac{a}{\left(x+1\right)\left(x+2\right)}+\frac{b}{\left(x+2\right)}=1\)

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

\(=\frac{\left(c+a\right)}{x^2+\left(2+x+1+\frac{a}{x^2+3ax+2a}+\frac{b}{x+2b}\right)=1}\)

\(=\frac{c+a}{x^2+\left(2c+3a+b\right)}x+2a+2b=0\)

\(\frac{c+a=0}{2c+3b=0}2a+2b=0\)

\(c=b=-a\)

Vậy:.....