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Question

$\dfrac{4}{x+2}+\dfrac{3}{2-x}+\dfrac{12}{x^2-4}$

Ngô Hải Nam · Accepted Answer

$\dfrac{4}{x+2}+\dfrac{3}{2-x}+\dfrac{12}{x^2-4}$$=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}$$=\dfrac{4x-8-3x-6+12}{\left(x+2\right)\left(x-2\right)}$$=\dfrac{4x-3x-8-6+12}{\left(x+2\right)\left(x-2\right)}$$=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}$ Câu trả lời: $\dfrac{4}{x+2}+\dfrac{3}{2-x}+\dfrac{12}{x^2-4}$$=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}$$=\dfrac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}$$=\dfrac{4x-8-3x-6+12}{\left(x+2\right)\left(x-2\right)}$$=\dfrac{4x-3x-8-6+12}{\left(x+2\right)\left(x-2\right)}$$=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}$

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