Câu hỏi của lộc Nguyễn

Question

ai onl giỏi toán 8 ko pm mình mik nhờ với

Minh Triều · Accepted Answer

$E=\frac{x^3+3x^2-4}{x^3-3x+2}$$=\frac{x^3+2x^2+x^2-4}{x^3-4x+x+2}$$=\frac{x^2.\left(x+2\right)+\left(x-2\right)\left(x+2\right)}{x.\left(x^2-4\right)+\left(x+2\right)}$$=\frac{\left(x+2\right)\left(x^2+x-2\right)}{x.\left(x-2\right)\left(x+2\right)+\left(x+2\right)}$$=\frac{\left(x+2\right)\left(x^2-x+2x-2\right)}{\left(x+2\right)\left[x.\left(x-2\right)+1\right]}$$\frac{\left(x+2\right)\left[x.\left(x-1\right)+2.\left(x-1\right)\right]}{\left(x+2\right)\left(x^2-2x+1\right)}$$=\frac{\left(x+2\right)\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)^2}=\frac{x+2}{x-1}$Câu trả lời: $E=\frac{x^3+3x^2-4}{x^3-3x+2}$$=\frac{x^3+2x^2+x^2-4}{x^3-4x+x+2}$$=\frac{x^2.\left(x+2\right)+\left(x-2\right)\left(x+2\right)}{x.\left(x^2-4\right)+\left(x+2\right)}$$=\frac{\left(x+2\right)\left(x^2+x-2\right)}{x.\left(x-2\right)\left(x+2\right)+\left(x+2\right)}$$=\frac{\left(x+2\right)\left(x^2-x+2x-2\right)}{\left(x+2\right)\left[x.\left(x-2\right)+1\right]}$$\frac{\left(x+2\right)\left[x.\left(x-1\right)+2.\left(x-1\right)\right]}{\left(x+2\right)\left(x^2-2x+1\right)}$$=\frac{\left(x+2\right)\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)^2}=\frac{x+2}{x-1}$

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