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+)\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{8x}{2\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}==\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}=\frac{x-2}{2\left(x+2\right)}\)
+) \(\frac{2x+6}{3x^2-x}-\frac{x+3}{1-3x}=\frac{2x+6}{x\left(3x-1\right)}-\frac{-x-3}{3x-1}\)
\(=\frac{2x+6}{x\left(3x-1\right)}-\frac{x\left(-x-3\right)}{x\left(3x-1\right)}\)
\(=\frac{2x+6+x^2+3x}{x\left(3x-1\right)}=\frac{x^2+5x+6}{x\left(3x-1\right)}\)
a)\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}\)
\(=\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2-8x}{2\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2+4x+4-8x}{2\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{2\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{2\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{2\left(x+2\right)}\)
b)\(\frac{2x+6}{3x^2-x}-\frac{x+3}{1-3x}\)
\(=\frac{2x+6}{x\left(3x-1\right)}-\frac{-x-3}{3x-1}\)
\(=\frac{2x+6+x\left(x+3\right)}{x\left(3x-1\right)}\)
\(=\frac{2x+6+x^2+3x}{x\left(3x-1\right)}\)
\(=\frac{x^2+5x+6}{x\left(3x-1\right)}\)
#Hoctot
\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3x^2+3x+3x+3}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3x\left(x+1\right)+3\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{\left(3x+3\right)\left(x+1\right)}\)
\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)^2}=\frac{x^2+1}{3\left(x+1\right)}\)
cho tam giác vuông ABC có E là trung điểm của AB , m là trung điểm của AC. chứng minh ABMN là hình j