nhanh hộ mình nhé.Mình đang gấp

x⁴+x²+1=x⁴+2x²+1-x²

=(x²+1)²-x²

=(x²-x+1)(x²+x+1)

1.

a. x³ - 4x² - xy² + 4x

= x ( x² - 4x + 4 - y² )

= x [ ( x - 2 )² - y² ]

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

b. x² - x - 2 = x² + x - 2x - 2 = x ( x + 1 ) - 2 ( x + 1 ) = ( x - 2 ) ( x + 1 )

c. x⁴ + 4

= ( x⁴ + 2x³ + 2x² ) - ( 2x³ + 4x² + 4x ) + ( 2x² + 4x + 4 )

= x² ( x² + 2x + 2 ) - 2x ( x² + 2x + 2 ) + 2 ( x² + 2x + 2 )

= ( x² + 2x + 2 ) ( x² - 2x + 2 )

Lời giải:

$3(x^4+x^2+1)(x^2+x+1)^2$

$=3[(x^4+2x^2+1)-x^2](x^2+x+1)^2$

$=3[(x^2+1)^2-x^2](x^2+x+1)^2$

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

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

a) \(8x^3-y^3-6xy\left(2x-y\right)=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2-6xy\right)=\left(2x-y\right)\left(4x^2-4xy+y^2\right)\)

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

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

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

a) 8x³ - y³ - 6xy(2x - y)

= (2x)³ - y³ - 3.2x.y.(2x - y)

= (2x - y)³

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

= (3x + 2 - x + 1)²

= (2x + 3)²

\(x^3+x^2+x+1\)

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

=\(x^2\left(x+1\right)+\left(x+1\right)\)=(x²+1)*(x+1)

Tích cho mình nhé,đúng đấy, không sai tí náo đâu

=(x⁴-1)+(x⁴-x²)

=(x²-1)(x²+1)+x²(x²-1)

=(x²-1)(2x²+1)

=(x-1)(x+1)(2x²+1)

= (x⁴-1) + (x⁴-x²)

= (x²-1)(x²+1)+x²(x²-1)  

= (x²-1)(2x²+1) 

= (x-1)(x+1)(2x²+1)                    

Mình cững đồng ý với bạn

Ta có :
\(x^2\left(x^4-1\right)\left(x^2+1\right)+1=x^2\left(x^2-1\right)\left(x^2+1\right)\left(x^2+2\right)+1\) 
\(\Leftrightarrow x^2\left(x^2+1\right)\left(x^2-1\right)\left(x^2+2\right)+1=\left(x^4-x^2\right)\left(x^4+x^2-2\right)+1\)
Gọi \(x^4-x^2\) là t, ta có:
t(t-2)+1=\(t^2-2t+1=\left(t-1\right)^2=\left(x^4+x^2-1\right)^2\)