a) 2x - 7 = 15 - (2 - 3x)

=> 2x - 7 = 15 - 2 + 3x

=> 2x - 7 = 13 + 3x

=> 2x - 3x = 13 + 7

=> -1x = 20

=> x = -20 thuộc Z

vậy______

b) 8 - ( x + 2 ) + ( 2x - 1 ) = 15 - ( 24 - 85 )

=> 8 - x - 2 + 2x - 1 = 15 - (-61)

=> (-x + 2x) + 8 - 2 - 1 = 76

=> 1x + 5 = 76

=> x = 76 - 5

=> x = 71 thuộc Z

vậy_____

c) -( 4x + 7 ) - ( 5 - 2x ) = 13 - x

=> -4x - 7 - 5 + 2x = 13 - x

=> (-4x + 2x) - 7 - 5 = 13 - x

=> -2x - 12 = 13 - x

=> -2x + x = 13 + 12

=> -1x = 25

=> x = - 25 thuộc Z

vậy____

a) 2x -7 = 15 - (2- 3x)

  2x - 7 = 15 - 2 +3x

  2x -3x = 15 - 2 + 7

  x( 2- 3) = 20

  -1x       = 20 

  x          = - 20

b) 8 - (x + 2) + (2x - 1) = 15 - (24 - 85)

  8 - x - 2 + 2x - 1 = 76

  (8 - 2 - 1) - (x - 2x) = 76

  5 - x(1 - 2) = 76

  5 - (-1x) = 76

  1x         = 76 -5

  1x          = 71

  x            = 71

2x + 15 = 4x - 3x + 8

2x - 4x + 3x = 8 - 15

x = -7

2x - 15 = 4x - 3x + 8

2x - x = 8 + 15

x = 23

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

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

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a) 2x+8≤ 0 

⇔2x≤-8

⇔x≤-4

b) 4x-7 ≥ 2x -5

⇔2x-12 ≥ 0

⇔2x≥12

⇔x≥6

c) (2x-8)(15-3x)>0

TH1: 2x-8>0 ⇒x>4

        15-3x>0⇒x<5 

TH2:  2x-8<0 ⇒x<4

        15-3x<0⇒x>5 (vô lí)

vậy 4<x<5

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

a) ( x + 2 )( 3 - 4x ) = x² + 4x + 4

<=> ( x + 2 )( 3 - 4x ) = ( x + 2 )²

<=> 3 - 4x = x + 2

<=> -4x - x = 2 - 3

<=> -5x = -1

<=> x = \(\frac{1}{5}\)

b) x(2x - 7) - 4x + 14 = 0

<=> x(2x - 7) = 4x - 14

<=> x(2x - 7) = 2(2x - 7)

<=> x = 2

c) 3x - 15 = 2x(x - 5)

<=> 3(x - 5) = 2x(x - 5)

<=> 3 = 2x

<=> x = \(\frac{3}{2}\)

d) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)

<=> 3x - 2 = 5x - 8

<=> 3x - 5x = -8 + 2

<=> -2x = -6

<=> x = 3

ôi A.R.M.Y

a)=3xy²

b) x⁵+4x³-6x²  : 4x²

     x⁵                : \(\overline{\frac{1}{4}x^3+x-1}\)

        4x³-6x²     :

        4x³           :

              -6x²    :

              -6x²     :

                 0    

Chào A.R.M.Y