Question

Cho biểu thức :

A=(x-1/x+2 - 2-5x/x^2-4) : x+3/x-2 (x khác +-2,x khác -3)

1) Rút gọn A

2)Tính giá trị của biểu thức A khi x =9

3)Tìm x để A<1

(Lm ơn giúp tui vs nha:3, tui đang cần gấp í )

Answer 1

`A=((x-1)/(x+2) - (2-5x)/(x^2-4)) : (x+3)/(x-2)`

`a, A = ((x-1)(x-2))/((x+2)(x-2)) - (2-5x)/((x-2)(x+2)) ) xx (x-2)/(x+3)`

`= ((x^2 - 3x + 2)/(x^2 -4) - (2-5x)/(x^2-4)) xx (x-2)/(x+3)`

`= (x^2 - 3x + 2 - 2 + 5x)/(x^2 - 4) xx (x-2)/(x+3)`

`= (x^2 + 2x)/((x-2)(x+2)) xx (x-2)/(x+3)`

`= (x(x+2))/((x-2)(x+2)) xx (x-2)/(x+3)`

`= x/(x+3)`

`b, x = 9 => A = 9/(9+3) = 9/12 = 3/4`

`c, A < 1 => x/(x+3) < 1 => (x+3-3)/(x+3) < 1 => 1 - 3/(x+3) < 1`

`=> x + 3 > 0`

`=> x > -3`

 

Answer 2

\(1,A=\left(\dfrac{\left(x-1\right)\left(x-2\right)-2+5x}{x^2-4}\right):\dfrac{x+3}{x-2}\\ =\dfrac{x^2-x-2x+2-2+5x}{x^2-4}.\dfrac{x-2}{x+3}\\ =\dfrac{x^2+2x}{x^2-4}.\dfrac{x-2}{x+3}=\dfrac{x\left(x+2\right)}{\left(x+2\right)}.\dfrac{1}{x+3}=\dfrac{x}{x+3}\)

\(2,x=9\left(thoanmanđk\right)\\ Thayx=9\\ \dfrac{9}{9+3}=\dfrac{9}{12}=\dfrac{3}{4}\)

c, \(đểA< 1\\\)

Xét hiệu A -1 <0

\(\dfrac{x}{x+3}-1< 0\\ \dfrac{x-x-3}{x+3}< 0\\ -\dfrac{3}{x+3}< \)

Ta có -3 < 0 

=> \(x+3< 0\\ x< -3\)