Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{x+2}=\frac{4-2x}{x+2}\)
\(ĐKXĐ:x\ne\pm2;x\ne0\)
\(A=\left(\frac{2}{2+x}-\frac{4}{x^2+4x +4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(A=\frac{4-2x}{x+2}\)
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn
Mình thử nha :33
ĐKXĐ : \(x\ne-3,x\ne-26,x\ne-6,x\ne1\)
Ta có :
\(A=\left[\frac{3}{2}-\left(\frac{x^4\left(x^2+1\right)-x^4-1}{x^2+1}\right)\cdot\frac{x^3-4x^2+\left(x-4\right)}{x^6\left(x+6\right)-\left(x+6\right)}\right]:\frac{\left(x+3\right)\left(x+26\right)}{3\left(x-2\right)\left(x+6\right)}\)
\(=\left[\frac{3}{2}-\left(\frac{x^6-1}{x^2+1}\right)\cdot\frac{\left(x-4\right)\left(x^2+1\right)}{\left(x+6\right)\left(x^6-1\right)}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)
\(=\left[\frac{3}{2}-\frac{x-4}{x+6}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)
\(=\frac{x+26}{2\left(x+6\right)}\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)
\(=\frac{3\left(x-2\right)}{2\left(x+3\right)}\)
Vậy : \(A=\frac{3\left(x-2\right)}{2\left(x+3\right)}\left(x\ne-3,x\ne-26,x\ne-6,x\ne1\right)\)
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(\text{Đ}K\text{X}\text{Đ}:x\ne\pm2\)
Ta có: \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\div\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2x+2-4}{\left(x+2\right)^2}\right):\left(\frac{2-x-2}{\left(x+2\right)\left(x-2\right)}\right)=\frac{2x-2}{\left(x+2\right)^2}\cdot\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{2\left(x-1\right)\left(x-2\right)}{-x\left(x+2\right)}\)