x⁸ + 2500

= (x⁴)² + 2.x⁴.50 + (50)² - 2.x⁴.50

= (x⁴ + 50)² - (10x²)²

= (x⁴ + 50 - 10x²)(x⁴ + 50 + 10x²)

Mình nghĩ là đề thiếu đó bạn :)

đề đáng lẽ phải là: \(x^7+x^2+1\)

\(x^7+x^2+1=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

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

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

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

Vay chac minh dua nam de XD

Bài khó quá

4x⁴+4x-3= (2x)²+2(2x)+1-4

=(2x+1)²-2²=(2x+1-2)(2x+1+2)

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

bn làm đ rùi, tui đ cho bn

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)

\(64-x^2-y^2+2xy=64-\left(x^2-2xy+y^2\right)\)

\(=8^2-\left(x-y\right)^2=\left(8-x+y\right)\left(8+x-y\right)\)

−(x−y−8)(x−y+8)

(x² + x)² - 2(x² + x) - 15

= [(x² + x)² - 2(x² + x) + 1] - 16

= (x² + x + 1)² - 4²

= (x² + x + 5)(x² + x - 3)

( x² + x )² - 2 ( x² + x ) - 15

Đặt t = x² + x , đa thức trở thành

t² - 2t - 15

= ( t² + 3t ) - ( 5t + 15 )

= t ( t + 3 ) - 5 ( t + 3 )

= ( t - 5 ) ( t + 3 )

= ( x² + x - 5 ) ( x² + x + 3 )

Đặt x² + x = t

bthuc <=> t² - 2t - 15

            = t² + 3t - 5t - 15

            = t( t + 3 ) - 5( t + 3 )

            = ( t + 3 )( t - 5 )

            = ( x² + x + 3 )( x² + x - 5 )

\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2