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Question

Rút gọn A = $\dfrac{3x}{x-1}$+$\dfrac{2}{x+1}$+$\dfrac{3-3x-2x^2}{x^2-1}$

Thanh Hoàng Thanh · Accepted Answer

$A=\dfrac{3x}{x-1}+\dfrac{2}{x+1}+\dfrac{3-3x-2x^2}{x^2-1}.$ $\left(ĐKXĐ:x\ne1;x\ne-1\right).$$A=\dfrac{3x\left(x+1\right)+2\left(x-1\right)+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.$$A=\dfrac{3x^2+3x+2x-2+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.$$A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}.$ Câu trả lời: $A=\dfrac{3x}{x-1}+\dfrac{2}{x+1}+\dfrac{3-3x-2x^2}{x^2-1}.$ $\left(ĐKXĐ:x\ne1;x\ne-1\right).$$A=\dfrac{3x\left(x+1\right)+2\left(x-1\right)+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.$$A=\dfrac{3x^2+3x+2x-2+3-3x-2x^2}{\left(x-1\right)\left(x+1\right)}.$$A=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}.$

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