x^4 - 4x^3 - 8x^2 - 16x + 16 
= x^4-8x^2+16-4x^3-16x 
= ( x^2+4)^2 - 4x(x^2+4 ) 
= ( x^2 + 4 )(x^2 + 4 - 4x) 
= (x^2 + 4 )( x - 2 )^2

\(x^4-4x^3+8x^2-16x+16\)

=\(x^4-4x^3+4x^2-16x+16\)

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

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

( 4x - 8 ) . ( x² +  6 ) - ( 4x - 8 ) . ( x + 7 )  +  9 . ( 8 - 4x )

= (4x - 8 ) . ( x² +  6 ) - ( 4x - 8 ) . ( x + 7 )  -  9 . (4x - 8)

= (4x-8) ( x²+6 - x - 7 - 9) 

= 4(x-2)(x^2 -x - 10) 

tre trau vai

a)

DKXD x khac 5

b) 

DKXD deo can vi da xac dinh voi moi x

c)

DKXD x khac -5

4x² - 4x + 1 - ( x + 2 )² = 0

[( 2x )² - 4x + 1 ] - ( x + 2 )² = 0

( 2x - 1 )² - ( x + 2 )² = 0

( 2x - 1 - x - 2 ) ( 2x - 1 + x + 2 ) = 0

( x - 3 ) ( 3x + 1 ) = 0

\(\Rightarrow\hept{\begin{cases}x-3=0\\3x+1=0\end{cases}}\)

1) x - 3 = 0

x = 3

2) 3x + 1 = 0

3x = -1

x= \(-\frac{1}{3}\)

\(4x^2-4x+1-\left(x+2\right)^2=0\)

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

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

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=3\end{cases}}\)

( x + 1 )² - ( x - 3 )² = 0

( x + 1 - x + 3 ) ( x + 1 + x - 3 ) = 0

4 ( 2x - 2 ) = 0

4 x 2 ( x - 1 ) = 0

8 ( x - 1 ) = 0

x - 1 = 0

x = 1

\(\left(x+1\right)^2-\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x+1+x-3\right)\left(x+1-x+3\right)=0\)

\(\Leftrightarrow4\left(2x-2\right)=0\)

\(\Leftrightarrow2x-2=0\)

\(\Leftrightarrow x=1\)

a)5x.(x-2y)+2.(2y-x)² = 5x.(x-2y)+2.(x-2y)² = (x-2y)[ 5x+2(x-2y)] = (x-2y)( 5x + 2x - 4y) = (x-2y)(7x-4y)

b) tương tự như trước, do là (A-B)^2 = a^2 - 2ab + b^2 = b^2 - 2ab + a^2 = ( b-a)^2 

c)100x²-(x²+25)²= (10x)^2 -(x²+25)²= (10x + x^2 + 25)( 10x-x^2-25) = (x^2+10x+25)[-(x^2-10x+25)] =-1(x+5)^2  (x-5)^2 

d)x²-xz-9y²+3y²= ( từ để suy nghĩ đã -_-)

e)x³-x^2.5x+125 = 

câu e dấu gì ở giữa vậy