a) 

<=> 3x - 3 + x - 2 = 2x - 2 - x + 1

<=> 3x + x - 2x + x = -2 + 1 + 3 + 2

<=>    3x               = 4

<=>   x                  = 4/3

Các câu sau làm tương tự

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

<=>   \(3x-3+x-2=2x-2-x+1\)

<=>  \(4x-5=x-1\)

<=> \(3x=4\)

<=>  \(x=\frac{4}{3}\)

Vậy....

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

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

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

Lời giải:

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

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

Giải:

a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)  

     \(\dfrac{-5}{6}-x=\dfrac{1}{4}\)

               \(x=\dfrac{-5}{6}-\dfrac{1}{4}\) 

               \(x=\dfrac{-13}{12}\) 

b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\) 

    \(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\) 

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

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

             \(x-\dfrac{1}{3}=\dfrac{1}{3}\) 

                    \(x=\dfrac{1}{3}+\dfrac{1}{3}\) 

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

c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\) 

           \(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\) 

            \(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\) 

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\) 

d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\) 

\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\) 

            \(x.\dfrac{5}{6}=\dfrac{29}{8}\) 

                \(x=\dfrac{29}{8}:\dfrac{5}{6}\) 

                \(x=\dfrac{87}{20}\)