Question

--------------GIÚP EM VS ẠK------------

câu 1:cho bt

A=(\(\dfrac{3-x}{x+3}\) . \(\dfrac{x^2+6x+9}{^{ }x^2-9}\)) :\(\dfrac{3x^2}{^{ }x+3}\)

a, rút gọn A

b, tính gt của biểu thức A tại x=\(\dfrac{2}{3}\)

c, tìm gt của x để A<0

Answer 1

a) \(A=\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\right).\dfrac{x+3}{3x^2}\)

\(\Leftrightarrow A=\left(\dfrac{-\left(x-3\right)}{x+3}.\dfrac{\left(x+3\right)}{x-3}\right).\dfrac{x+3}{3x^2}\)

\(\Leftrightarrow A=\dfrac{-x-3}{3x^2}\)

b) Khi \(x=\dfrac{2}{3}\Rightarrow\) \(A=\dfrac{-\dfrac{2}{3}-3}{3.\left(\dfrac{2}{3}\right)^2}=\dfrac{-11}{4}\)

c) Để A < 0 thì

\(\dfrac{-x-3}{3x^2}< 0\)

=> -x -3 <0

<=> -x < 3

\(\Rightarrow x>3\)