Question

Cho biểu thức A= (\(\dfrac{x}{x^2+2x}+\dfrac{x}{x^2-2x}\)) (1-\(\dfrac{2}{x}\)) ( với x ≠ -2 , 2, 0)

a) Rút gọn A

b) Tìm x để |A|+A=0

Answer 1

a,A=(\(\dfrac{x}{x^2+2x}+\dfrac{x}{x^2-2x}\))(1-\(\dfrac{2}{x}\))

=\(\dfrac{x\left(x-2\right)+x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}.\dfrac{x-2}{x}\)

=\(\dfrac{x\left(x-2+x+2\right)}{x\left(x-2\right)\left(x+2\right)}.\dfrac{x-2}{x}\)

=\(\dfrac{2}{x+2}\)

Vậy.....

Answer 2

b, /A/\(\ge\)0 \(\forall\)x

Để /A/+A=0 thì A\(\le\)0

<=>\(\dfrac{2}{x+2}\le0\)

<=>x+2\(\le\)0

<=>x\(\le\)-2

Mà ĐKXĐ: x\(\ne\)-2

=> x<-2

Vậy để /A/+A=0 thì x<2