Question

Cho biểu thức \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

a) Rút gọn A

b) Tính giá trị biểu thức khi \(\left|x\right|=\frac{1}{2}\)

Answer 1

Lời giải:

a) ĐKXĐ: $x\neq 0; x\neq \pm 2$

\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)

\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)

b)

Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)

\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)

\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)

Answer 2

Lời giải:

a) ĐKXĐ: $x\neq 0; x\neq \pm 2$

\(A=\left(\frac{x^2}{x(x^2-4)}-\frac{6}{3(x-2)}+\frac{1}{x+2}\right):\frac{(x-2)(x+2)+10-x^2}{x+2}\)

\(=\left(\frac{x}{(x-2)(x+2)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2(x+2)+(x-2)}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{-6}{(x-2)(x+2)}.\frac{x+2}{6}=\frac{1}{2-x}\)

b)

Khi \(|x|=\frac{1}{2}\Rightarrow x=\pm \frac{1}{2}\) (thỏa mãn ĐKXĐ)

\(x=\frac{1}{2}\Rightarrow A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}\)

\(x=-\frac{1}{2}\Rightarrow A=\frac{1}{2--\frac{1}{2}}=\frac{2}{5}\)