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

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

\(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}=\dfrac{x-5}{\left(x-2\right)^2}.\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)

dễ mà bro 

a)x+5= 9

x=9-5

x=4

b) -3x=24

x=24:-3

x=-8

c) [x]-9=15

[x]=15+9

[x]=24

x=24

d)[x-9]=-2+17

[x-9]=15

[x]=15+9

[x]=24

x=24

a) \(x+5=-2+11\)                                b)\(-3x=5+29\)

\(\Rightarrow x=-2+11-5\)                                   \(\Rightarrow-3x=24\)

\(\Rightarrow x=4\)                                                       \(\Rightarrow-8\)        

c)\(\left|x\right|-9=-2+17\)                         d)\(\left|x-9\right|=-2+17\)

     \(\left|x\right|=15+9\)                                    \(\left|x-9\right|=15\)

      \(\left|x\right|=24\)                                          \(x-9=15\) hoặc \(x-9=15\)

x = 24 hoặc - 24                                    \(x=24\) hoặc \(x=-6\)

Bài 2: 

a: =>x-1=1 hoặc x-1=-1

=>x=2 hoặc x=0

b: =>x+1=-1

hay x=-2

c: =>(135-7x):9=8

=>135-7x=72

=>7x=63

hay x=9

d: =>(x+7)(x-3)<0

=>-7<x<3

e: \(\Leftrightarrow3^{x-3}=18+9=27\)

=>x-3=3

hay x=6

f: =>4-2x=0

hay x=2

ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)

a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

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

\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

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

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

\(=\dfrac{2x}{3x-y}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

\(=\dfrac{4x+5}{x.\left(x+5\right)}-\dfrac{3x}{x.\left(x+5\right)}\)

\(=\dfrac{x+5}{x.\left(x+5\right)}\)

\(=\dfrac{1}{x}\)

a) Ta có: \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

\(=\dfrac{x\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}+\dfrac{x\left(3x+y\right)}{\left(3x+y\right)\left(3x-y\right)}+\dfrac{2xy}{\left(3x+y\right)\left(3x-y\right)}\)

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

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

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

\(=\dfrac{2x}{3x-y}\)

b) Ta có: \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

\(=\dfrac{4x+5}{x\left(x+5\right)}-\dfrac{3x}{x\left(x+5\right)}\)

\(=\dfrac{4x+5-3x}{x\left(x+5\right)}\)

\(=\dfrac{x+5}{x\left(x+5\right)}\)

\(=\dfrac{1}{x}\)

a: \(x_1+x_2=-\dfrac{b}{a}=2\sqrt{3};x_1\cdot x_2=\dfrac{c}{a}=1\)

Đặt \(A=x_1-x_2\)

=>\(A^2=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)

\(=\left(2\sqrt{3}\right)^2-4\cdot1=12-4=8\)

=>\(\left[{}\begin{matrix}x_1-x_2=2\sqrt{2}\\x_1-x_2=-2\sqrt{2}\end{matrix}\right.\)

b: \(\dfrac{3x_1^2+5x_1x_2+3x_2^2}{4x_1^3\cdot x_2+4x_1\cdot x_2^3}\)

\(=\dfrac{3\left(x_1^2+x_2^2\right)+5x_1x_2}{4x_1x_2\left(x_1^2+x_2^2\right)}\)

\(=\dfrac{3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+5x_1x_2}{4x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)

\(=\dfrac{3\left(x_1+x_2\right)^2-x_1x_2}{4x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}\)

\(=\dfrac{3\cdot12-1}{4\cdot1\cdot\left[12-2\cdot1\right]}=\dfrac{35}{4\cdot10}=\dfrac{7}{8}\)