có nhìn thấy j đâu  ri msk

`@` `\text {Ans}`

`\downarrow`

`A= (2x - 3)^2 - (2x + 3)^2`

`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`

`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`

`= -6 * 4x`

`= -24x`

`A=(2x-3)^2-(2x+3)^2`

`A=(2x-3-2x-3)(2x-3+2x+3)`

`A=-6.4x=-24x`

b: B=(x+1)^2-2(2x-1)(x+1)+4x^2-4x+1

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

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

=(-x+2)^2=x^2-4x+4

dài quá, làm từ từ nhé

1, \(\left(a-b\right)^2\left(2a-3b\right)-\left(b-a\right)^2\left(3a-5b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=\left(a-b\right)^2\left(2a-3b-3a+5b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=\left(a-b\right)^2\left(-a+2b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=-\left(a-b\right)^2\left(a-2b\right)+\left(a+b\right)^2\left(a-2b\right)\)

\(=\left(a-2b\right)\left[\left(a+b\right)^2-\left(a-b\right)^2\right]\)

\(=\left(a-2b\right)\left(a+b-a+b\right)\left(a+b+a-b\right)\)

\(=4ab\left(a-2b\right)\)

2, \(x^4-4\left(x^2+5\right)-25=\left(x^2-25\right)-4\left(x^2+5\right)=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)

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

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

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

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

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

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

4, câu này đề thiếu

5,\(16\left(xy+6\right)^2-\left(4x^2+y^2-25\right)^2=\left(4xy+24\right)^2-\left(4x^2+y^2-25\right)^2\)

\(=\left(4xy+24-4x^2-y^2+25\right)\left(4xy+24+4x^2+y^2-25\right)\)

\(=\left[49-\left(4x^2-4xy+y^2\right)\right]\left[\left(4x^2+4xy+y^2\right)-1\right]\)

\(=\left[49-\left(2x-y\right)^2\right]\left[\left(2x+y\right)^2-1\right]\)

\(=\left(7-2x+y\right)\left(7+2x-y\right)\left(2x+y-1\right)\left(2x+y+1\right)\)

6, \(\left(x+y-2z\right)^2+\left(x+y+2z\right)^2-16z^2\)

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

\(=\left(x+y-2z\right)^2+\left(x+y-2z\right)\left(x+y+6z\right)\)

\(=\left(x+y-2z\right)\left(x+y-2z+x+y+6z\right)\)

\(=\left(x+y-2z\right)\left(2x+2y+4z\right)\)

\(=2\left(x+y-2z\right)\left(x+y+2z\right)\)

7,\(=a^2x^2+6axy+9y^2-\left(-6ax^2-6ay^2+x^2+y^2\right)+9x^2-6axy+a^2y^2\)

\(=a^2x^2+6axy+9y^2+6ax^2+6ay^2-x^2-y^2+9x^2-6axy+a^2y^2\)

\(=a^2x^2+6ax^2+8x^2+a^2y^2+6ay^2+8y^2\)\(=x^2\left(a^2+6a+8\right)+y^2\left(a^2+6a+8\right)\)

\(=\left(x^2+y^2\right)\left(a^2+6a+8\right)\)\(=\left(x^2+y^2\right)\left(a^2+2a+4a+8\right)\)

\(=\left(x^2+y^2\right)\left[a\left(a+2\right)+4\left(a+2\right)\right]=\left(x^2+y^2\right)\left(a+2\right)\left(a+4\right)\)

a) 4x - 1 = 3x - 2

<=> 4x - 3x = -2 + 1

<=> x = -1

Vậy S = {-1}

b) x + 1 = 2. (x - 3)

<=> x + 1 = 2x - 6

<=> x - 2x = -6 -1 

<=> -x = -7

<=> x = 7

Vậy:....

c) 2. (x + 1) + 3 = 2 - x

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

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

<=> 3x = -3

<=> x = -1

Vậy:,,,,

a,  4x-1=3x-2                                        b,              x+1=2(x-3)

    4x-3x= -2+1                                                      x+1=2x-6

     x      = -1                                                           x-2x=-6-1

   Vậy Tập nghiệm S =( -1 )                                   -x  = -7

                                                                                 x=7

                                                               Vậy tập nghiệm S =(7)

c,   2(x+1)+3=2-x

      2x+2+3=2-x

       2x+5    =2-x

       2x+x    =2-5

         3x      = -3

           x      -1

Vậy tập nghiệm S = (-1)                                                    

a) 4x -1 = 3x -2 

<=> 4x - 3x = - 2 + 1

<=> x = -1

Vậy tập nghiệm S ={-1}

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

<=> x+1 = 2x - 6

<=> 1 + 6 = 2x - x

<=> x = 7

Vậy tập nghiệm S ={7}

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

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

<=> 2x  + x = 2 - 2 - 3 

<=> 3x = -3

<=> x = -1

Vậy tập nghiệm S ={-1}

a, \(x^2+7x+10-12x+9=x^2-10x+25\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)

b, bạn ktra lại đề nhé 

c, \(x^2-4+3x+3=3+x^2-x-2\)

\(\Leftrightarrow x^2+3x-1=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

a, \(5a^2-5b^2=5\left(a^2-b^2\right)=5\left(a+b\right)\left(a-b\right)\)

b, \(2ab^2-a^2b-b^3=b\left(2ab-a^2b-b^2\right)=-b\left(a^2-2ab+b^2\right)\)

\(=-b\left(a-b\right)^2\)

a) 5a² - 5b² = 5(a²-b²) = ( a+b) (a-b)

b) 2ab² - a²b - b³ = b ( 2ab-a²-b² ) = -b (a²-2ab+b²) = -b (a-b)²

c: \(=\left(a-b\right)\left(a^2+ab+b^2\right)+ab\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+2ab+b^2\right)=\left(a-b\right)\left(a+b\right)^2\)

d: \(=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

e: \(=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)

a )

1. \(\sqrt{x^2-4x+3}=x-2\)

<=> x² - 4x + 3 = (x - 2)²

<=> x² - 4x + 3 = x² - 4x + 4

<=> x² - x² - 4x + 4x = 1

<=> 0 = 1 (Vô lí)

vậy PT có nghiệm là S = \(\varnothing\)

2. \(\sqrt{4x^2-4x+1}=x-1\)

<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)

<=> 2x - 1 = x - 1

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

<=> x = 0

1: ta có: \(\sqrt{x^2-4x+3}=x-2\)

\(\Leftrightarrow x^2-4x+3=x^2-4x+4\)(vô lý)

2: Ta có: \(\sqrt{4x^2-4x+1}=x-1\)

\(\Leftrightarrow\left(2x-1-x+1\right)\left(2x-1+x-1\right)=0\)

\(\Leftrightarrow x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)