[(2x)²-2.2x.3+3²]-2(x²-2x+x-2)=[(3x)²+2.3x.1+1²]+2(1-2x)

<=>4x²-12x+9-(2x²-4x+2x-4)=9x²+6x+1+2-4x

<=>4x²-2x²-9x²-12x+4x-2x-6x+4x+9+4-1-2=0

<=>-7x²-12+10=0 

đen ta = (-12)²-4(-7)10=424

phương trình có 2 ngiệm phân biệt 

x₁=

x₂=

1.\(x^2-2x-4y^2-4y=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)

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

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

4.\(x^3-4x^2-9x+36=x^2\left(x-4\right)-9\left(x-4\right)=\left(x-3\right)\left(x+3\right)\left(x-4\right)\)

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

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

7.\(\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)\(=\left(x-5\right)\left(4x+1\right)\)

8.\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)=\left(3x-2\right)\left(3x-6\right)=3\left(3x-2\right)\left(x-2\right)\)

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

=>\(6x^2+2x-6x-2-\left(6x^2-9x-4x+6\right)=5\)

=>\(6x^2-4x-2-6x^2+13x-6=5\)

=>9x-8=5

=>9x=13

=>\(x=\frac{13}{9}\)

2: \(\left(1-3x\right)\left(3x-5\right)-\left(2x-4\right)\left(2-3x\right)=x-6\)

=>\(3x-5-9x^2+15x+\left(2x-4\right)\left(3x-2\right)=x-6\)

=>\(-9x^2+18x-5+6x^2-4x-12x+8=x-6\)

=>\(-3x^2+2x+3-x+6=0\)

=>\(-3x^2+x+9=0\)

=>\(3x^2-x-9=0\)

=>\(x^2-\frac13x-3=0\)

=>\(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{109}{36}=0\)

=>\(\left(x-\frac16\right)^2=\frac{109}{36}\)

=>\(x-\frac16=\pm\frac{\sqrt{109}}{6}\)

=>\(x=\frac16\pm\frac{\sqrt{109}}{6}\)

3: \(\left(2x-1\right)\left(4x^2+2x+1\right)-\left(2x+1\right)\left(4x^2-2x+1\right)=5x+6\)

=>\(8x^3-1-8x^3-1=5x+6\)

=>5x+6=-2

=>5x=-8

=>\(x=-\frac85\)

\(2\left(x-2\right)=x\left(x-2\right)\)

\(\Rightarrow2\left(x-2\right)-x\left(x-2\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\2-x=0\end{cases}\Rightarrow x=2}\)

a)2

b)\(\frac{-11}{4}\)

b) mình khỏi ghi đề lại ha :3

=> 2x^2 - 4x + 2 + 3x^2 + 12x + 12 - 25x^2 + 1= 15

sau đó bạn gom lại những số vd như là 4x với 12x,..... rồi tính ra đc là

-20x^2 + 8x + 15 = 15

=> -20x^2 + 8x = 0

=> 2x ( -10x + 4 ) = 0

=> 2x = 0 => x= 0

hoặc -10x +4 = 0

        => -10x = -4

        => x     = 4/ 10

a) ( 2x-3)^ 2 - ( 2x + 5) ^ 2 = 18

=> 4x^2 - 12x + 9 - ( 4x^2 + 20x + 25 ) = 18

=>  4x^2 - 12x + 9 - 4x^2 - 20x - 25  = 18

=> (4x^2- 4x^2) + (-12x - 20x) + ( 9 -25 ) = 18 

=>         0         -          32x    -       16    = 18

=>          -32x                                       = 32

=>             x                                         = -1

bạn đợi mình type câu b :v

Nhầm 1 chút nhé, Bài 1 câu a) ( x^2 - 5x ) . ( x^2 - x +3) - ( x^3 - x^2 +1) . ( 2x - 1 )

a, 7/2 + 2x = 16/3 : 8/3

  7/2 + 2x   = 2

     2x          = 2 - 7/2

        2x       = -3/2

          x       = -3/2 : 2

          x       =  -3/4

b (13/3 + 3x) * 13/5 = -13/3

   13/3 + 3x               = -13/3 : 13/5

   13/3 + 3x               = -5/3

     3x                         = -5/3 - 13/3

      3x                        = -6

         x                      = -6 : 3

         x                      = -2 

mình ko viết lại đề bài nha.

i8uAywq2ruihUIEHTUIQW4E