a)    15-(49-4x)=3x-27

<=> 15-49+4x=3x-27

<=> -34+4x-3x=-27

<=> x=-27+34=7

Vậy x=7

b)   70-(x-37)-(-61)=14-2x

<=> 70-x+34+61=14-2x

<=> 120-x+2x=14

<=> x=120-14=106

Vậy x=106

1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)

2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)

3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)

\(=7x^2-24x+13\)

4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)

\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)

5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)

6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)

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

\(\dfrac{x-1}{13}-\dfrac{2x-13}{15}=\dfrac{3x-15}{27}-\dfrac{2x-27}{29}\)

\(\Leftrightarrow\dfrac{x-1}{13}-1-\dfrac{2x-13}{15}-1=\dfrac{3x-15}{27}-1-\dfrac{2x-27}{29}-1\)

\(\Leftrightarrow\dfrac{x-1-13}{13}-\dfrac{2x-13-15}{15}=\dfrac{3x-15-27}{27}-\dfrac{4x-27-29}{29}\)

\(\Leftrightarrow\dfrac{x-14}{13}-\dfrac{2x-24}{15}=\dfrac{3x-42}{27}-\dfrac{4x-56}{29}\)

\(\Leftrightarrow\dfrac{x-14}{13}-\dfrac{2\left(x-14\right)}{15}-\dfrac{3\left(x-14\right)}{27}-\dfrac{4\left(x-14\right)}{29}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}-\dfrac{4}{29}\right)=0\)

\(\Leftrightarrow x-14=0\) ( Vì: \(\dfrac{1}{13}-\dfrac{2}{15}-\dfrac{3}{27}-\dfrac{4}{29}\ne0\))

\(\Leftrightarrow x=14\)

1) 5x = 120 : 2

5x = 60

x = 60 : 5

x = 12.

2) 3x + 27 = 45

3x = 45 - 27

3x = 18

x = 18 : 3

x = 6.

3) 73 + x - 122 = 234 : 2

73 + x - 122 = 117

73 + x = 117 + 122

73 + x = 239

x = 239 - 73

x = 166.

4) 32 - 2 ( x - 1 ) = 12

2 ( x - 1 ) = 32 - 12

2 ( x - 1 ) = 20

x - 1 = 20 : 2

x - 1 = 10

x = 10 + 1

x = 11.

5) 3 ( x + 1 ) - 15 = 36

3 ( x + 1 ) = 36 + 15

3 ( x + 1 ) = 51

x + 1 = 51 : 3

x + 1 = 17

x = 17 - 1

x = 16.

6) 32 - ( 4x - 16 ) = 21

4x - 16 = 32 - 21

4x - 16 = 11

4x = 11 + 16

4x = 27

x = 27 : 4

x = 6,75.

7) 13x - 3x = 500

( 13 - 3 )x = 500

10x = 500

x = 500 : 10

x = 50.

8) 9x + x - 35 = 68

9x + 1x - 35 = 68

9x + 1x = 68 + 35

9x + 1x = 103

( 9 + 1 )x = 103

10x = 103

x = 103 : 10

x = 10,3.

1) 5x = 120 : 2

5x = 60

x = 60 : 5

x = 12.

2) 3x + 27 = 45

3x = 45 - 27

3x = 18

x = 18 : 3

x = 6.

3) 73 + x - 122 = 234 : 2

73 + x - 122 = 117

73 + x = 117 + 122

73 + x = 239

x = 239 - 73

x = 166.

4) 32 - 2 ( x - 1 ) = 12

2 ( x - 1 ) = 32 - 12

2 ( x - 1 ) = 20

x - 1 = 20 : 2

x - 1 = 10

x = 10 + 1

x = 11.

5) 3 ( x + 1 ) - 15 = 36

3 ( x + 1 ) = 36 + 15

3 ( x + 1 ) = 51

x + 1 = 51 : 3

x + 1 = 17

x = 17 - 1

x = 16.

6) 32 - ( 4x - 16 ) = 21

4x - 16 = 32 - 21

4x - 16 = 11

4x = 11 + 16

4x = 27

x = 27 : 4

x = 6,75.

7) 13x - 3x = 500

( 13 - 3 )x = 500

10x = 500

x = 500 : 10

x = 50.

8) 9x + x - 35 = 68

9x + 1x - 35 = 68

9x + 1x = 68 + 35

9x + 1x = 103

( 9 + 1 )x = 103

10x = 103

x = 103 : 10

x = 10,3.

a/ \(2x+\frac{1}{7}=\frac{1}{3}\)

=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)

=> \(2x=\frac{4}{21}\)

=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)

b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)

=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)

=> \(x-\frac{1}{2}=\frac{4}{27}\)

=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)

c/ \(\left(x-5\right)^2+4=68\)

=> \(\left(x-5\right)^2=68-4=64\)

=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)

d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)

=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)

e) \(5x+2=3x+8\)

=> \(5x-3x=8-2=6\)

=> \(2x=6\)

=> \(x=6:2=3\)

f/ \(26-\left(5-2x\right)=27\)

=> \(5-2x=26-27=-1\)

=> \(2x=5-\left(-1\right)=5+1=6\)

=> \(x=6:2=3\)

g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)

=> \(4x-8-2x+6=4\)

=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)

=> \(2x+-2=4\)

=> \(2x=4+2=6\)

=> \(x=6:2=3\)

h/ \(\left(x+3\right)^3:3-1=-10\)

=> \(\left(x+3\right)^3:3=-10+1=-9\)

=> \(\left(x+3\right)^3=-9.3=-27\)

=> \(x+3=-3\)

=> \(x=-3-3=-6\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1