Question

[nimbaoi12xyz]

giúp em với ạ e cảm ơn !!

---

Answer 1

[Đoàn Đức Hà]

Điều kiện: \(x\ne1,x\ne0\).

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

\(=\left(\frac{x^2-2x+1}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right)\div\frac{2}{x^2+1}\)

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

\(=\frac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{2}\)

\(=\frac{x^3-1}{x^3-1}.\frac{x^2+1}{2}=\frac{x^2+1}{2}\)

\(4E=2\left(x^2+1\right)=x+8\)

\(\Leftrightarrow2x^2-x-6=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)(thỏa mãn) 

\(E=\frac{x^2+1}{2}>0\)do \(x^2\ge0\).