\(1,\frac{x+1}{x-2}=\frac{3}{4}\)

\(\Rightarrow3x-6=4x+4\)

\(\Rightarrow3x-4x=4+6\)

\(\Rightarrow-x=10\Leftrightarrow x=-10\)

\(2,\frac{x-1}{3}=\frac{x+3}{5}\)

\(\Rightarrow5x-5=3x+9\)

\(\Rightarrow5x-3x=9+5\)

\(\Rightarrow2x=14\Leftrightarrow x=7\)

\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)

\(\Rightarrow64x+96=72x-24\)

\(\Rightarrow72x-64x=24+96\)

\(\Rightarrow8x=120\)

\(\Rightarrow x=15\)

x/2 = 3/4

1. \(\frac{x-3}{x+2}=\frac{3}{4}\)

=> \(4(x-3)=3(x+2)\)

=> 4x - 12 = 3x + 6

=> 4x - 12 - 3x = 6

=> 4x - 3x - 12 = 6

=> x = 18

2. \(\frac{4-x}{x-1}=\frac{-5}{6}\)

=> 6[4 - x] = -5[x - 1]

=> 24 - 6x = 5 - 5x

=> 24 - 6x - 5 = 5x

=> 24 - 5 = 6x - 5x

=> 19 = x

=> x = 19

3. \(\frac{3-x}{-x+2}=\frac{-1}{3}\)

=> 3[3 - x] = -1[-x + 2]

=> 9 - 3x = x - 2

=> 9 - 3x + x = -2

=> 9 - 4x = -2

=> 4x = 9 - [ -2]

=> 4x = 9 + 2 = 11

=> x = 11/4

4. \(\frac{x-1}{12}=\frac{-3}{1-x}\)

=> [x-1][1-x] = -36

=> -[x-1]² = -36

=> [x-1]² = 36

=> [x-1]² = 6²

=> x -1 = 6 hoặc x - 1 = -6

=> x = 7 hoặc x = -5

5 . |x - 3 | - 8 = -4

=> |x - 3| = -4 + 8

=> |x - 3| = 4

=> x - 3 = 4 hoặc x - 3 = -4

=> x = 7 hoặc x = -1

Mấy bài sau tương tự

1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

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

\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

Ta có : 6x² - 11x + 3 

= 6x² - 2x - 9x + 3

= (6x² - 2x) - (9x - 3)

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

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

K MIK NHA BẠN !!!!!!!!!!

bÀI 1 

bÀI 2 : 

Bài 3 :

Bài 4: 

5,

6, 

7, 

8,

9, 

10,

11,

12,

13,

K MIK NHA BẠN !!!!!!!!!!

Nguyễn Quang Trung Linh Nhi : Mik cảm ơm 2 bạn nhiều nhóe ;) 

-3-2x+17=12-3x

-2x+3x=12-17+3

x=-2

b.43-2x-2=-48+x

  -2x-x = -48+2-43

   -3x=-89

    x=-89/-3

c  -253-2x=102+19-x

        -2x+x=102+19+253

         -x=374

             x=-374

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

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

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

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

a,-0,162

\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)

Vậy nghiệm của phương trình trên là \(2\)

\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình trên là \(-2\)

\(c.\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\left(x\ne2\right)\\ \Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\\\Leftrightarrow \frac{2\left(x+5\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\\\Leftrightarrow 2\left(x+5\right)-3\left(x-2\right)=3\left(2x-3\right)\\\Leftrightarrow 2x+10-3x+6=6x-9\\\Leftrightarrow 2x-3x-6x=-10-6-9\\\Leftrightarrow -7x=-25\\\Leftrightarrow x=\frac{25}{7}\left(tm\right)\)

Vậy nghiệm của phương trình trên là \(\frac{25}{7}\)