Ta có:

3 - x . 9 - x 2 = 3 - x 3 - x 3 + x = 3 - x 2 1)

Và 3 + x x 2 - 6 x + 9 = 3 + x . x - 3 2 = 3 + x . 3 - x 2 (2)

( vì ( x- 3) = - (3- x) nên x - 3 2 = - 3 - x 2 = 3 - x 2  )

Từ (1) và (2) suy ra: x - 3 . 9 - x 2 = 3 + x x 2 - 6 x + 9

Do đó: 

\(\frac{x^2\left(x+2\right)}{x\left(x+2\right)^2}=\frac{x}{x+2}\Rightarrow\frac{x}{x+2}=\frac{x}{x+2}\)

\(\frac{3-x}{3+x}=\frac{x^2-6x+9}{9-x^2}\Rightarrow\frac{3-x}{3+x}=\frac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}\Rightarrow\frac{3-x}{3+x}=\frac{3-x}{3+x}\)

\(\frac{x^3-4x}{10-5x}=\frac{-x^2-2x}{5}\Rightarrow-\frac{x\left(x-2\right)\left(x+2\right)}{5\left(x-2\right)}=\frac{-x^2-2x}{5}\)

\(\Rightarrow\frac{-x\left(x+2\right)}{5}=\frac{-x^2-2x}{5}\Rightarrow\frac{-x^2-2x}{5}=\frac{-x^2-2x}{5}\)

k nha bạn

sai rồi cái này là dùng định nghĩa 2 phân thức bằng nhau để chứng minh chúng bằng nhau mà

Bài 1: 

c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

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

Ta có:

x 2 x + 2 x + 2 = x 2 x + 2 2 x x + 2 2 . x = x 2 x + 2 2 S u y   r a :   x 2 x + 2 x + 2 = x 2 x + 2 2

vậy

a. \(x^2y^3.35xy=5.7x^3y^4\)

\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)

\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)

\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)

\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)

\(\Leftrightarrow\left(3-x\right)\left(3-x\right)\left(3+x\right)=\left(3+x\right)\left(3-x\right)^2\)

\(\Leftrightarrow\left(3-x\right)^2\left(3+x\right)=\left(3-x\right)^2\left(3+x\right)\)

\(\Rightarrowđpcm\)

\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)

\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)

Ta có: x 3 - 4 x . 5 = 5 x 3 - 20 x

10 - 5 x - x 2 - 2 x = - 10 x 2 - 20 x + 5 x 3 + 10 x 2 = 5 x 3 - 20 x

Suy ra: x 3 - 4 x . 5 = 10 - 5 x - x 2 - 2 x

Vậy