\(16-x^2\)

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

\(---\)

\(16-3x+1^2\) (kt lại đề bài nhé)

\(x^4y^4+4x^2y^2+4\)

\(=\left[\left(xy\right)^2\right]^2+2\cdot\left(xy\right)^2\cdot2+2^2\)

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

\(---\)

\(y^2-4y+4-x^2\)

\(=y^2-2\cdot y\cdot2+2^2-x^2\)

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

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

\(\sqrt{9.\left(x-1\right)^2}-12=0\)

=> 3.(x - 1) - 12 = 0

=> 3x - 15 = 0

=> 3x = 15

=> x = 5

b) \(\sqrt{4.\left(3-x\right)}=16\) (ĐKXĐ: x ≤ 3)

\(\Rightarrow\sqrt{3-x}=8\)

=> 3 - x = 64

=> x = -61

1: =(16x^2-8x+1)-y^2

=(4x-1)^2-y^2

=(4x-1-y)(4x-1+y)

2: =(x^2-2xy+y^2)-z^2

=(x-y)^2-z^2

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

3: =(x^2+4xy+4y^2)-16

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

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

4: =(x^2-4xy+4y^2)-16

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

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

\(9\left(x-1\right)^2-16\left(x-2\right)^2\)

\(=\left(3x-3\right)^2-\left(4x-8\right)^2\)

\(=\left(3x-3-4x+8\right)\left(3x-3+4x-8\right)\)

\(=\left(-x+5\right)\left(7x-11\right)\)

16) 2x + 2y - x² - xy = ( 2x + 2y ) - ( x² + xy ) = 2( x + y ) - x( x + y ) = ( x + y )( 2 - x )

17) x² - 2x - 4y² - 4y = ( x² - 4y² ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2( x + 2y ) = ( x + 2y )( x - 2y - 2 )

18) x²y - x³ - 9y + 9x = ( x²y - x³ ) - ( 9y - 9x ) = x²( y - x ) - 9( y - x ) = ( y - x )( x² - 9 ) = ( y - x )( x - 3 )( x + 3 )

19) x²( x - 1 ) + 16( 1 - x ) = x²( x - 1 ) - 16( x - 1 ) = ( x - 1 )( x² - 16 ) = ( x - 1 )( x - 4 )( x + 4 )

20) 2x² + 3x - 2xy - 3y = ( 2x² - 2xy ) + ( 3x - 3y ) = 2x( x - y ) + 3( x - y ) = ( x - y )( 2x + 3 )

20, \(2x^2+3x-2xy-3y=2x\left(x-y\right)+3\left(x-y\right)=\left(2x+3\right)\left(x-y\right)\)

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

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

18, \(x^2y-x^3-9y+9x=-x\left(x^2-9\right)+y\left(x^2-9\right)=\left(-x-y\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)

19, \(x^2\left(x-1\right)+16\left(1-x\right)=x^2\left(x-1\right)-16\left(x-1\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)

Bài 1: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

16) 2x + 2y - x² - xy

   =  ( 2x - x² ) + ( 2y - xy )

   =   x ( 2 - x )  +  y ( 2 - x ) 

   =   ( 2 - x ) ( x + y )

17) x² - 2x - 4y² - 4y

    = ( x² - 4y² ) - ( 2x + 4y )

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

    = ( x + 2y ) ( x - 2y - 2 )

18) x²y - x³ - 9y +9x

    = ( 9x + x³ ) + ( x²y - 9y )

    =  x ( 9 + x² ) +  y ( x² - 9 ) 

    =   x ( 9 + x² ) - y ( 9 + x² )

    = ( 9 + x² ) ( x - y )

    = ( 3 - x ) ( 3 + x ) ( x - y )

19) x² ( x - 1) + 16 (1 - x )

 =   x² ( x - 1 ) - 16 ( x - 1 )

 = ( x - 1 ) ( x² - 16 )

 = ( x - 1 ) ( x - 4 ) ( x + 4 )

20) 2x² + 3x - 2xy - 3y

   =  2x² + 3x - ( 2xy + 3y )

   =   x ( 2x + 3 ) - y ( 2x + 3 )

   = ( 2x + 3 )  ( x - y )

\(4^{2-x}=4^{16x}\)

\(2-x=16x\)

\(17x=-2\)

\(x=-\dfrac{2}{17}\)