Question

\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}+\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{2x+1}{x+3}-1\right)\)

jup mik với mik cần gấp tính P=? nha

Answer 1

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`

`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`

`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`

`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`

`=(2x+6)/(x-2)`

Answer 2

ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)

\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)

\(=\dfrac{\left(3+x\right)^2-\left(3-x\right)^2-4x^2}{\left(3+x\right)\left(3-x\right)}:\dfrac{x-2}{x+3}\)

\(=\dfrac{\left(3+x-3+x\right)\left(3+x+3-x\right)-4x^2}{\left(x+3\right)\left(3-x\right)}.\dfrac{x+3}{x-2}\)

\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)

\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)

 

Answer 3

Chắc mình làm tắt quá để mình làm lại bước biến đổi.

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((x-3)/(x+3)+(4x^2)/(x^2-9)-(x+3)/(x-3)):((2x+1-x-3)/(x+3))`

`=((x-3)^2/(x^2-9)+(4x^2)/(x^2-9)-(x+3)^2/(x^2-9)):((x-2)/(x+3))`

`=(((x-3)^2+4x^2-(x+3)^2)/(x^2-9))*(x+3)/(x-2)`

`=(x^2-6x+9+4x^2-x^2-6x-9)/(x^2-9)*(x+3)/(x-2)`

`=(4x^2-12x)/(x^2-9)*(x+3)/(x-2)`

`=(4x(x-3))/((x-3)(x+3))*(x+3)/(x-2)`

`=(4x)/(x+3)*(x+3)/(x-2)`

`=(4x)/(x-2)`

Answer 4

P = \(\left(\dfrac{x+3}{3-x}-\dfrac{3-x}{x+3}-\dfrac{4x^2}{9-x^2}\right):\dfrac{2x+1-x-3}{x+3}\)

=  \(\left[\dfrac{\left(x+3\right)^2-\left(x-3\right)^2-4x^2}{\left(3-x\right)\left(x+3\right)}\right]:\dfrac{x-2}{x+3}\)

= \(\dfrac{x^2+6x+9-x^2+6x-9-4x^2}{\left(3-x\right)\left(x+3\right)}.\dfrac{x+3}{x-2}\)

= \(\dfrac{-4x^2+12x}{\left(3-x\right)\left(x+3\right)}.\dfrac{x+3}{x-2}\)

= \(\dfrac{4x\left(3-x\right)}{\left(3-x\right)\left(x-2\right)}=\dfrac{4x}{x-2}\)