4x^4 - 32x^2 +1 = 4x^4 + 4x^2 +1 - 36x^2 = (2x^2 + 1)^2 - 36x^2 = (2x^2 - 6x + 1)(2x^2 + 6x + 1) 

4 x⁴ - 32 x² + 1 

= ( 2 x² )² - 2 . 2x². 8 + 64 - 63 

= ( 2 x² - 8 )² - 63 

= ( 2x² - 8 + √63 ) ( 2x² - 8 - √63 )

Xong

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

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

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

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

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

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

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

b,   x²-2xy+y²-z²

=(x-y)²-z²

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

con bai 32, 33 neu ban tra loi duoc minh h them

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

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

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

x^5+x+1

=x(x^4+1)+1 

=(x^2+x+1)(x^3-x^2+1) 

Ta có : x⁵ + x + 1 

= x⁵ + x⁴ - x⁴ - x³ + x³ + x² - x² - x + x + 1

=  (x⁵ + x⁴) - (x⁴ + x³) + (x³ + x²) - (x² + x) + (x + 1)

= x⁵(x + 1) - x⁴.(x + 1) + x³(x + 1) - x²(x + 1) + (x + 1)

= (x + 1)(x⁵ - x⁴ + x³ - x² + 1)

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

bn có phải Mod Yonnii bên hoidap247 ko ak?

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

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)

hc tốt nha !!!!!!!!!

KHÔNG BÍT

4x²-4x+1=(2x)²-2.2x+1²=(2x-1)²

\(4x^2-4x+1\)\(=\) \((2x^2)-2.2x.1+1^2\)

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

Học tốt!