Question

Cho A = \(\frac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\)

a, Rút gọn

b, Tìm x để A =0

Answer 1

Có: A=\(\frac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\)

A=\(\frac{\left(x^3-4x^2\right)+\left(3x^2-10x-8\right)}{x^2\left(x-4\right)+5\left(x-4\right)}\)

A=\(\frac{x^2\left(x-4\right)+\left(3x^2-12x+2x-8\right)}{\left(x^2+5\right)\left(x-4\right)}\)

A=\(\frac{x^2\left(x-4\right)+3x\left(x-4\right)+2\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) ĐKXĐ:\(x\ne4\)

A=\(\frac{\left(x^2+3x+2\right)\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left(x^2+x+2x+2\right)\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left[x\left(x+1\right)+2\left(x+1\right)\right]\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left(x+1\right)\left(x+2\right)\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}\)

Vậy A=\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}\)với \(x\ne4\)

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Answer 2

b) Có A=\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}\text{với x}\ne4\)

A=0⇔\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}=0\)

⇔(x+1)(x+2)=0 (vì \(x^2+5\ne0\))

⇔\(\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)(Thoả mãn ĐKXĐ)

Vậy với x=1 hoặc x=2 thì A=0