\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

a) Rút gọn thu được kết quả: 3;

b) Ta có MC = 3x (x - 3)

Thực hiện tính toán thu được kết quả: x 2 − 6 x + 9 3 x ( x − 3 ) = x − 3 3 x  

c) Trước tiên biến đổi: 3 + 3 x = 3 ( x + 1 ) x ; 3 3 ( x + 1 ) x = x x + 1  

Thay vào A và thu gọn ta được A = 4 x + 3 x

giúp tui với mn ơi

\(=x^2-9-x^2=-9\)

\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

\(\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=\dfrac{0}{x\left(x-3\right)}=0\)

\(\left(x^3-x^2-5x-3\right):\left(x-3\right)\\ =\left[\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(x-3\right)\right]:\left(x-3\right)\\ =\left[x^2\left(x-3\right)+2x\left(x-3\right)+\left(x-3\right)\right]:\left(x-3\right)\\ =\left[\left(x-3\right)\left(x^2+2x+1\right)\right]:\left(x-3\right)\\ =x^2+2x+1\)

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

`@` `\text {Ans}`

`\downarrow`

`(x+3)(x-4)`

`= x(x-4) + 3(x-4)`

`= x^2-4x + 3x - 12`

`= x^2 - x - 12`

=x^2-4x+3x-12

=x^2-x-12

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

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