Question

[lộc Nguyễn]

ai onl giỏi toán 8 ko pm mình mik nhờ với

Answer 1

[Minh Triều]

\(E=\frac{x^3+3x^2-4}{x^3-3x+2}\)

\(=\frac{x^3+2x^2+x^2-4}{x^3-4x+x+2}\)

\(=\frac{x^2.\left(x+2\right)+\left(x-2\right)\left(x+2\right)}{x.\left(x^2-4\right)+\left(x+2\right)}\)

\(=\frac{\left(x+2\right)\left(x^2+x-2\right)}{x.\left(x-2\right)\left(x+2\right)+\left(x+2\right)}\)

\(=\frac{\left(x+2\right)\left(x^2-x+2x-2\right)}{\left(x+2\right)\left[x.\left(x-2\right)+1\right]}\)

\(\frac{\left(x+2\right)\left[x.\left(x-1\right)+2.\left(x-1\right)\right]}{\left(x+2\right)\left(x^2-2x+1\right)}\)

\(=\frac{\left(x+2\right)\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)^2}=\frac{x+2}{x-1}\)