`@` `\text {Ans}`

`\downarrow`

`x^2 + 4x + 3`

`= x^2 + 3x + x + 3`

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

`= x(x + 3) + (x + 3)`

`= (x+1)(x+3)`

____

`2x^2 + 3x - 5`

`= 2x^2 + 5x - 2x - 5`

`= (2x^2 - 2x) + (5x - 5)`

`= 2x(x - 1) + 5(x - 1)`

`= (2x + 5)(x - 1)`

____

`16x - 5x^2 - 3`

`= 15x + x - 5x^2 - 3`

`= (15x - 5x^2) + (x - 3)`

`= 5x(3 - x) + (x - 3)`

`= -5x(x - 3) + (x - 3)`

`= (1 - 5x)(x - 3)`

\(x^2+4x+3=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\\----\\ 2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\\ ----\\ 16x-5x^2-3=-5x^2+15x+x-3=-5x\left(x+3\right)-\left(x+3\right)=-\left(5x+1\right)\left(x+3\right)\)

\(x^2+4x+3=\left(x+1\right)\left(x+3\right)\\ 2x^2+3x-5=\left(2x+5\right)\left(x-1\right)\\ 16x-5x^2-3=\left(1-5x\right)\left(x-3\right)\)

Bài 1 : 

a, \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)

\(=x^2+6x+9+x^2-6x+9+2x^2-18\)

\(=4x^2\)

b, \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9=8\)

Bài 2 : 

a, \(16x-8xy+xy^2=x\left(16-8y+y^2\right)=x\left(4-y\right)^2\)

b, \(3\left(3-x\right)-2x\left(x-3\right)=3\left(3-x\right)+2x\left(3-x\right)=\left(3+2x\right)\left(3-x\right)\)

c, \(3x^2+4x-4=3x^2+6x-2x-4=\left(x+2\right)\left(3x-2\right)\)

Rút gọn biểu thức

a) ( x + 3 )²  + ( x - 3 )² + 2( x² - 9 )

= x³ + 6x + 9 + x² - 6x + 9 + 2x² - 18

= 4x²

b) ( 4x - 1 )³ - ( 4x - 3 )( 16x² + 3 )

= 64x³ - 48x² + 12x - 1 - ( 64x³ - 48x² + 12x - 9 )

= 64x³ - 48x² + 12x - 1 - 64x³ + 48x² - 12x + 9

= 8

PTĐTTNT

a) 16x - 8xy + xy²

= x( 16 - 8y + y² )

= x( 4 - y )²

b) 3( 3 - x ) ± 2x( x - 3 ) < không biết thay dấu gì (: >

= 3( 3 - x ) \(\mp\)2x( 3 - x )

= ( 3 - x )( 3 \(\mp\)2x ) 

c) 3x² + 4x - 4

= 3x² + 6x - 2x - 4

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

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

Tìm x

a) ( 3x - 2 )( 3x + 4 ) - ( 2 - 3x )² = 6

<=> ( 3x - 2 )( 3x + 4 ) - ( 3x - 2 )² = 6

<=> ( 3x - 2 )( 3x + 4 - 3x + 2 ) = 6

<=> ( 3x - 2 ).6 = 6

<=> 3x - 2 = 1

<=> x = 1

b) 2( x - 3 ) - ( x - 3 )( 3x - 2 ) = 0

<=> ( x - 3 )( 2 - 3x + 2 ) = 0

<=> ( x - 3 )( 4 - 3x ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{3}\end{cases}}\)

c) ( x - 1 )( x + 2 ) - x( x - 2 ) = -5

<=> x² + x - 2 - x² + 2x = -5

<=> 3x - 2 = -5

<=> 3x = -3

<=> x = -1

a.5x²-10xy+5y²-20z²

  =5(x²-2xy+y²-4z²)

  =5[ (x²-2xy+y²)-(2z)² ]

  =5[ (x-y)²-(2z)² ]

  =5(x-y-2z)(x-y+2z)

b.16x-5x²-3

  =15x+x-5x²-3

  =(15x-3)+(x-5x²)

  =3(5x-1)+x(1-5x)

  =3(5x-1)-x(5x-1)

  =(5x-1)(3-x)

c.x²-5x+5y-y²

  =(5y-5x)+(x²-y²)

  =5(y-x)+(x-y)(x+y)

  =5(y-x)-(y-x)(y+x)

  =(y-x)[5-(y+x)]

  =(y-x)(5-y-x)

d.3x²-6xy+3y²-12z²     (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)

=3(x²-2xy+y²-4z²)

=3[ (x²-2xy+y²)-(2z)² ]

=3[ (x-y)²-(2z)² ]

=3(x-y-2z)(x-y+2z)

e.x²+4x+3

=x²+3x+x+3

=(x²+x)+(3x+3)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f.(x²+1)²-4x²

=(x²+1)²-(2x)²

=(x²+1-2x)(x²+1+2x)

h.x²-4x-5

=x²-5x+x-5

=(x²+x)+(-5x-5)

=x(x+1)-5(x+1)

-(x+1)(x-5)

\(a,3x^3-6x^2+3x\)

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

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

\(b,16x^2y-4xy^2-4x^3\)

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

\(=-4x\left(x-2y+y\sqrt{3}\right)\left(x-2y-y\sqrt{3}\right)\)

nhiều quá, các bn ngại làm, chia nhỏ ra,mk làm cho 2 câu 

a) x² +5x -6 = x² -x +x + 5x -6

= x² -x +6x -6

= x( x-1) + 6(x-1) = (x-1)(x+6)

b) 5x² +5xy -x-y = 5x(x+y) -(x+y)

= (x+y)(5x-1)

e) \(2x^2+3x-5\)

\(=2x^2+5x-2x-5\)

\(=x\cdot\left(2x+5\right)-\left(2x+5\right)\)

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

f) \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15+x-3\)

\(=-5\cdot\left(x-3\right)+x-3\)

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

c)7x-\(6x^2\)-2

=3x + 4x - \(6x^2\) - 2

=(3x - \(6x^2\)) - (2 - 4x)

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

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

a,\(x^2+4x+3\)

=\(x^2+3x+x+3\)

=\(x\left(x+3\right)+\left(x+3\right)\)

=(x+3)(x+1)

b,\(2x^2+3x-5\)

=\(2x^2+5x-2x-5\)

=x(2x+5)-(2x+5)

=(2x+5)(x-1)

c,\(16x-5x^2-3\)

=\(-\left(5x^2-16x+3\right)\)

=\(-\left(5x^2-x-15x+3\right)\)

=-[x(5x-1)-3(5x-1)]

=-[(5x-1)(x-3)]

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

\(a.\) \(x^2+4x+3\)

\(=x^2+x+3x+3\)

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

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

\(b.\) \(2x^2+3x-5\)

\(=2x^2-2x+5x-5\)

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

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

\(c.\)\(16x-5x^2-3\)

\(=-5x^2+16x-3\)

\(=-5x^2+15x+x-3\)

\(=-5x\left(x-3\right)+\left(x-3\right)\)

\(=\left(x-3\right)\left(1-5x\right)\)

bài 1:

a. x² - 5=0

=>x² = 0+5 = 5

=> x = \(\sqrt{5}\)

vậy x= \(\sqrt{5}\)

sorry biết mỗi a thôi

a) x² - 5 = 0

    x²       = 0 + 5

    x²        = 5

=> x = \(\sqrt{5}\)

Vậy ...

\(x^2-5=0\)

<=>  \(\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)

<=>  \(\orbr{\begin{cases}x-\sqrt{5}=0\\x+\sqrt{5}=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

Bài 1.

\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)

\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)

a) \(4x\left(a-b\right)+6xy\left(b-a\right)\)

\(=4x\left(a-b\right)-6xy\left(a-b\right)\)

\(=\left(4x-6xy\right)\left(a-b\right)\)

\(=2x\left(2-3y\right)\left(a-b\right)\)

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

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

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

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

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

g: \(\left(3x-1\right)^2-\left(x+3\right)^2\)

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

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

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