Câu hỏi của Gia huy

Question

C=3x^2-x+2/(x-1) (x+3) + x/x-1 - x-1/x+3Giúp mình gấp với

Tran Le Khanh Linh · Accepted Answer

$C=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x}{x-1}-\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)$$=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}$$=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-1\right)\left(x+3\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}$$=\frac{3x^2-x+2-x^2-3x-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}$$=\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}=\frac{x-1}{x+3}$Vậy C=$\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)$Câu trả lời: $C=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x}{x-1}-\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)$$=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}$$=\frac{3x^2-x+2}{\left(x-1\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-1\right)\left(x+3\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}$$=\frac{3x^2-x+2-x^2-3x-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}$$=\frac{x^2-2x+1}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}=\frac{x-1}{x+3}$Vậy C=$\frac{x-1}{x+3}\left(x\ne1;x\ne-3\right)$

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