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ĐKXĐ : \(x\ne\pm2;x\ne0;x\ne3\)
\(A=\left(\frac{4x}{2+x}+\frac{8x^2}{4-x^2}\right):\left(\frac{x-1}{x^2-2x}-\frac{2}{x}\right)\)
\(=\frac{4x\left(2-x\right)+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\frac{8x-4x^2+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}:\frac{3-x}{x\left(x-2\right)}\)
\(=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(x-2\right)}{3-x}\) \(=\frac{4x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(=\frac{4x^2}{x-3}\)
\(A< 0\Leftrightarrow\frac{4x^2}{x-3}< 0\Leftrightarrow x-3< 0\) ( do \(4x^2>0\) )
\(\Leftrightarrow x< 3\)
Vậy :........
\(DKXD:x\ne\pm2;x\ne3;x\ne\frac{3}{2};x\ne0\)
\(A=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-3x}\right)\)
\(=\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x^2-3x}{x^2-3x}\)
\(=\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{x\left(2x-3\right)}{x\left(x-3\right)}\)
\(=\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2x-3}{x-3}\)
\(=\frac{4x\left(2x-3\right)}{\left(2+x\right)\left(x-3\right)}\)
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 .....
\(B=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{x^2-3x}{2x^2-x^3}\left(ĐKXĐ:x\ne2;-2;0\right)\)
a)\(B=\left(-\frac{\left(x+2\right)^2}{x^2-4}-\frac{4x^2}{x^2-4}+\frac{\left(x-2\right)^2}{x^2-4}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(B=\left(\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{x^2-4}\right).\frac{-x\left(x-2\right)}{\left(x-3\right)}\)
\(B=\left(\frac{-x^2-4x-4-4x^2+x-4x+4}{\left(x-2\right)\left(x+2\right)}\right).-\frac{x\left(x-2\right)}{x-3}\)
\(B=\frac{-5x^2-7x}{\left(x+2\right)}.\frac{-x}{x-3}\)
\(B=\frac{\left(-5x^2-7x\right)-x}{\left(x+2\right)\left(x-3\right)}\)
\(B=\frac{5x^3+7x^2}{\left(x+2\right)\left(x+3\right)}\)
Hình như đề sai.Sửa đề luôn nha !
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{-6}=\frac{1}{x-2}\)
b
Để \(A< 0\Rightarrow\frac{1}{x-2}< 0\Rightarrow x-2< 0\Rightarrow x< 2\)
c
Để A nguyên thì \(\frac{1}{x-2}\) nguyên
\(\Rightarrow1⋮x-2\)
\(\Rightarrow x-2\in\left\{1;-1\right\}\Rightarrow x\in\left\{3;1\right\}\)
ĐKXĐ: \(x\ne\pm2;x\ne0\)
\(A=\left[\frac{4x\left(x-2\right)}{x^2-4}-\frac{8x^2}{x^2-4}\right]:\left[\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right]\)
\(=\frac{-4x^2-8x}{x^2-4}:\frac{-x+3}{x\left(x-2\right)}\)
\(=\frac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}.\frac{x\left(x-2\right)}{-x+3}\)
\(=\frac{4x^2}{x-3}\)
Vì \(4x^2\ge0\)với mọi x nên:
để A > 0 thì x - 3 >0 <=> x > 3