e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

b) \(-x^2-12x+21=\left(3-x\right)\left(x+11\right).\)

\(\Leftrightarrow-x^2-12x+21=-x^2-8x+33\)

\(\Leftrightarrow33+4x=21\)

\(\Leftrightarrow-4x=12\)

\(\Rightarrow x=-3\)

c,\(9x+5x^2+1=5x^2-22+13x\)

\(\Leftrightarrow4x-22=1\)

\(\Leftrightarrow4x=23\)

\(\Rightarrow x=\frac{23}{4}\)

Mk làm mẫu cho 1 pt nha !

a, 

pt <=> 4x^2-7x+5 = 2x^2-5x-18 

<=> (4x^2-7x+5)-(2x^2-5x-18) = 0

<=> 4x^2-7x+5-2x^2+5x+18 = 0

<=> 2x^2-2x+23 = 0

<=> x^2-x+23/2 = 0

<=> (x^2-x+1/4)+45/4 = 0

<=> (x-1/2)^2+45/4 = 0

=> pt vô nghiệm [ vì (x-1/2)^2+45/4 > 0 ]

P/S: Tham khảo nha

a) Trường hợp 1. Xét 4 - 5x = 5 - 6x.

Tìm được x = 1.

a: =>(x-3)(x+1)=0

=>x=3 hoặc x=-1

b: =>x(x-3)=0

=>x=0 hoặc x=3

c: =>(x-5)(x+1)=0

=>x=5 hoặc x=-1

d: =>5x^2+7x-5x-7=0

=>(5x+7)(x-1)=0

=>x=1 hoặc x=-7/5

e: =>x^2-4=0

=>x=2 hoặc x=-4

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

=>(x-2)^2=3

=>\(x=2\pm\sqrt{3}\)

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

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

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

\(\Leftrightarrow2x^2+2-2x^2-2x=0\)

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

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

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)

a) Ta có: \(x^2-2x-3=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: \(S_1=\left\{3;-1\right\}\)(1)

Ta có: \(\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy: \(S_2=\left\{-3;-1\right\}\)(2)

Từ (1) và (2) suy ra \(S_1\ne S_2\)

hay Hai phương trình \(x^2-2x-3=0\) và \(\left(x+1\right)\left(x+3\right)=0\) không tương đương với nhau

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

a , 2x -3 = 5x + 6

    2x -5x=6+3

    -3x = 9

     x =9 :(-3)

   x= -3

a) 2x-5x=3+6

-3x=9

x=-3

vậy........

b)(2x+1).(3x-2)-(5x-8).(2x+1)=0

(2x+1).(3x-2-2x-1)=0

(2x-1).(x-3)=0

==>x=1/2 ; x=3

c)(2x+1).5-(7x+5)=(2x-2).3

10x+5-7x-5=6x-6

3x=6x-6

3x-6x=6

-3x=6

x=-2

a) 2x - 3 = 5x + 6

<=> -3x = 9

<=> x = -3

b) (2x + 1).(3x - 2) = (5x - 8).(2x + 1)

<=> 6x² - 4x + 3x - 2 = 10x² + 5x - 16x -8

<=> -4x² - 10x + 6 = 0

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

c) \(\frac{2\text{x}+1}{3}-\frac{7\text{x}+5}{15}=\frac{2\text{x}-2}{5}\)

<=> \(\frac{5.\left(2\text{x}+1\right)}{5.3}-\frac{7\text{x}+5}{15}=\frac{3.\left(2\text{x}-2\right)}{3.5}\)

<=> 10x + 5 - 7x + 5 - 6x + 6 = 0

<=> -3x + 16 = 0

<=> -3x = -16

<=> x = \(\frac{16}{3}\)

d) \(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> \(\frac{3\text{x}\left(x-5\right)-x\left(x-2\right)}{\left(x-2\right).\left(5-x\right)}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> 3x² - 15x - x² + 2x - 3x = 0

*Câu e dễ quá, bạn tự làm nhé :v*