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\(A=\)\(\frac{x|x-2|}{x^2+8x-20}+12x-3.\)
\(=\frac{x|x-2|}{\left(x-2\right)\left(x+10\right)}+12x-3\)
Nếu \(x\ge2\Rightarrow x-2\ge0\Leftrightarrow|x-2|=x-2\)
\(\Rightarrow A=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\frac{x}{x+10}+12x-3\)
Nếu \(x< 2\Rightarrow x-2< 0\Leftrightarrow|x-2|=-\left(x-2\right)\)
\(\Rightarrow A=\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3=\frac{-x}{x+10}+12x-3\)
Ta có:\(\frac{\left[x\left(x-2\right)\right]}{x^2+8x-20}+12x-3=\frac{x\left(x-2\right)}{x^2-2x+10x-20}+12x-3\)
\(=\frac{x\left(x-2\right)}{x\left(x-2\right)+10\left(x-2\right)}+12x-3=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}+12x-3\)
\(=\frac{x}{x+10}+12x-3=\frac{x+\left(12x-3\right).\left(x+10\right)}{x+10}=\frac{x+12x^2+120x-3x-30}{x+10}\)
\(=\frac{12x^2+118x-30}{x+10}\)