Question

cho biểu thức:

A= (\(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\)) : \(\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)

a) Rút gọn A

b) tìm x để A < 1

Answer 1

\(ĐKXĐ:x\ne0;x\ne\pm3\)

\(A=\left(\frac{1}{3}+\frac{3}{x\left(x-3\right)}\right):\left(\frac{x^2}{3\left(9-x^2\right)}+\frac{1}{x+3}\right)\\ =\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{-3\left(x+3\right)\left(x-3\right)}{x^2-3x+9}\\ =\frac{-x-3}{x}\)

b) Ta có :

\(A=\frac{-x-3}{x}< 1\\ \Leftrightarrow\frac{-x-3}{x}-1< 0\\ \Leftrightarrow\frac{-x}{x}-\frac{3}{x}-1< 0\\ \Leftrightarrow-1-1-\frac{3}{x}< 0\\ \Leftrightarrow-2-\frac{3}{x}< 0\\ \Leftrightarrow\frac{-3}{x}< 2\\ \Leftrightarrow2x< -3\\ \Rightarrow x>\frac{-3}{2}=-1,5\)

Vậy để A < 1 thì x > 1,5 / x ≠ 0 ; x ≠ 3 ; x ≠ -3