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\(\text{Đ}K\text{X}\text{Đ}:x\ne\pm2\)
Ta có: \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\div\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left(\frac{2x+2-4}{\left(x+2\right)^2}\right):\left(\frac{2-x-2}{\left(x+2\right)\left(x-2\right)}\right)=\frac{2x-2}{\left(x+2\right)^2}\cdot\frac{\left(x+2\right)\left(x-2\right)}{-x}\)
\(=\frac{2\left(x-1\right)\left(x-2\right)}{-x\left(x+2\right)}\)
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
A = \(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right).\left(x+1\right)}-\frac{x+3}{2\left(x+2\right)}\right).\frac{4x^2-4}{5}\)
A = \(\left(\frac{\left(x+1\right)^2+3.2-\left(x+3\right).\left(x-1\right)}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4x^2-4}{5}\)
A = \(\left(\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right).\left(x+1\right)}\right).\frac{4\left(x^2-1\right)}{5}\)
A = \(\frac{10}{2\left(x-1\right).\left(x+1\right)}.\frac{4\left(x-1\right).\left(x+1\right)}{5}\)
A = 4
\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)
\(A=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-3x}\right)\)
\(=\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x^2-3x}{x^2-3x}\)
\(=\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2x-3\right)}{x\left(x-3\right)}\)
\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)
\(=\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}\)
b
Xét hơi bị nhiều TH nhá:(
Để \(A>0\) thì \(\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}>0\)
TH1:\(4x\left(2x-3\right)>0;\left(2+x\right)\left(x-3\right)>0\)
\(TH2:4x\left(2x-3\right)< 0;\left(2+x\right)\left(x-3\right)< 0\)
Bạn tự xét nốt nhá!
c
\(\left|x-7\right|=4\Rightarrow x-7=4;x-7=-4\)
\(\Rightarrow x=11;x=3\)
Thay vào .....
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{x+2}=\frac{4-2x}{x+2}\)
\(ĐKXĐ:x\ne\pm2;x\ne0\)
\(A=\left(\frac{2}{2+x}-\frac{4}{x^2+4x +4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(A=\frac{4-2x}{x+2}\)