Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

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

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

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

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

\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

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

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

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

bài 3 A) \(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

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

x=0

x=2

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

a) \(x^2-81=\left(x-9\right)\left(x+9\right)\)

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

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

d) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

e) \(6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)

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

g) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

h) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

Bài 3

a) x² + 10x + 25

= x² + 2.x.5 + 5²

= (x + 5)²

b) 8x - 16 - x²

= -(x² - 8x + 16)

= -(x² - 2.x.4 + 4²)

= -(x - 4)²

c) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

d) (x + y)² - 9x²

= (x + y)² - (3x)²

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

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

e) (x + 5)² - (2x - 1)²

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

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

Bài 4

a) x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

b) (x - 4)² - 36 = 0

(x - 4 - 6)(x - 4 + 6) = 0

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

x - 10 = 0 hoặc x + 2 = 0

*) x - 10 = 0

x = 10

*) x + 2 = 0

x = -2

Vậy x = -2; x = 10

c) x² - 10x = -25

x² - 10x + 25 = 0

(x - 5)² = 0

x - 5 = 0

x = 5

d) x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

(x² + 2x) + (3x + 6) = 0

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

x + 2 = 0 hoặc x + 3 = 0

*) x + 2 = 0

x = -2

*) x + 3 = 0

x = -3

Vậy x = -3; x = -2