d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)

\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow-x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)

\(\Leftrightarrow3x-3-3=2x+6\)

\(\Leftrightarrow x=12\)

b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)

\(\Leftrightarrow3x+12-4x-12=2x+12\)

\(\Leftrightarrow-3x=12\)

hay x=-4

c: Ta có: \(\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Bài 1:

a. 

$(4x^2+4x+1)-x^2=0$

$\Leftrightarrow (2x+1)^2-x^2=0$

$\Leftrightarrow (2x+1-x)(2x+1+x)=0$

$\Leftrightarrow (x+1)(3x+1)=0$

$\Rightarrow x+1=0$ hoặc $3x+1=0$

$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$

b.

$x^2-2x+1=4$

$\Leftrightarrow (x-1)^2=2^2$

$\Leftrightarrow (x-1)^2-2^2=0$

$\Leftrightarrow (x-1-2)(x-1+2)=0$

$\Leftrightarrow (x-3)(x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-1$

c.

$x^2-5x+6=0$

$\Leftrightarrow (x^2-2x)-(3x-6)=0$

$\Leftrightarrow x(x-2)-3(x-2)=0$

$\Leftrightarrow (x-2)(x-3)=0$

$\Leftrightarrow x-2=0$ hoặc $x-3=0$

$\Leftrightarrow x=2$ hoặc $x=3$

2c.

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$

$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$

$\Leftrightarrow x=-4$ (tm)

2d.

ĐKXĐ: $x\neq 2$

PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$

$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$

$\Rightarrow 3x-5=3-x$

$\Leftrightarrow 4x=8$

$\Leftrightarrow x=2$ (không tm) 

Vậy pt vô nghiệm.

2f.

ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$

$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$

$\Leftrightarrow x^2-4x+4-3x-6=2x-22$

$\Leftrightarrow x^2-7x-2=2x-22$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Leftrightarrow x-4=0$ hoặc $x-5=0$

$\Leftrightarrow x=4$ hoặc $x=5$ (tm)

a: Ta có: \(3x-\left(3x+2\right)=x+3\)

\(\Leftrightarrow x+3=-2\)

hay x=-5

b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)

\(\Leftrightarrow15x-3+8x-4=18x\)

\(\Leftrightarrow5x=7\)

hay \(x=\dfrac{7}{5}\)

\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)

\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)

`<=> 20(2x-1) +15(3x-2) =12(4x-3)`

`<=> 40x - 20 + 45x - 30 = 48x - 36`

`<=> 85x -50 = 48x - 36`

`<=> 85x-48x = -36+50`

`<=> 37x =14`

`<=> x= 14/37`

Vậy phương trình có nghiệm `x=14/37`

__

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)

\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

`=> 5x + 15 + 4x -12=x-6`

`<=> 9x + 3=x-6`

`<=> 9x-x=-6-3`

`<=> 8x = -9`

`<=>x=-9/8(tm)`

Vậy phương trình có nghiệm `x=-9/8`

` @ yngoc`

a: =>-3x=-12

=>x=4

b: =>3(3x+2)-3x-1=12x+10

=>9x+6-3x-1=12x+10

=>12x+10=6x+5

=>6x=-5

=>x=-5/6

c: =>x(x+1)+x(x-3)=4x

=>x^2+x+x^2-3x-4x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=3(loại) hoặc x=0(nhận)

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

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

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

a, 4x+1=13-2x <-->6x=12 <-->x=2

b, (2x-5)(x-4)=0 <-->x=5/2  hoặc x=4

c,Đề bài -->x(x-2)+6(x+2)=2x+12 -->x^2+2x=0 -->x=0  hoặc x=-2

d,|x-3|=9-2x -->TH1: x-3=9-2x -->x=x=4           TH2:3-x=9-2x -->x=6