Question

Cho biểu thức C= \(\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)

a) Rút gọn
b)tìm x thuộc Z để C đạt giá trị nguyên dương
C) Tìm x để \(C^2-C+1\)

Answer 1

a: \(C=\left(\dfrac{2x^2+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\dfrac{x^2+x+1-x^2+2}{x^2+x+1}\)

\(=\dfrac{2x^2+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+3}\)

\(=\dfrac{x^2-x}{\left(x-1\right)}\cdot\dfrac{1}{x+3}=\dfrac{x}{x+3}\)

b: Để C là số nguyên dương thì \(\left\{{}\begin{matrix}x⋮x+3\\\dfrac{x}{x+3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3\in\left\{1;-1;3;-3\right\}\\x\in\left(-\infty;-3\right)\cup\left(0;+\infty\right)\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{-4;-6\right\}\)