a) x3 + y3 - 3xy + 1
= ( x + y )3 - 3xy( x + y ) - 3xy + 1
= [ ( x + y )3 + 1 ] - [ 3xy( x + y ) + 3xy ]
= ( x + y + 1 )( x2 + 2xy + y2 - x - y + 1 ) - 3xy( x + y + 1 )
= ( x + y + 1 )( x2 - xy + y2 - x - y + 1 )
b) ( 4 - x )5 + ( x - 2 )5 - 32
= [ -( x - 4 ) ]5 + ( x - 2 )5 - 32
Đặt t = x - 3
đthức <=> ( 1 - t )5 + ( 1 + t )5 - 32 ( chỗ này bạn dùng nhị thức Newton để khai triển nhé )
= 10t4 + 20t2 - 30
Đặt y = t2
đthức = 10y2 + 20y - 30
= 10y2 - 10y + 30y - 30
= 10y( y - 1 ) + 30( y - 1 )
= 10( y - 1 )( y + 3 )
= 10( t2 - 1 )( t2 + 3 )
= 10( t - 1 )( t + 1 )( t2 + 3 )
= 10( x - 3 - 1 )( x - 3 + 1 )[ ( x - 3 )2 + 3 ]
= 10( x - 4 )( x - 2 )( x2 - 6x + 12 )
a,\(x^3+y^3-3xy+1\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy\)
\(=\left[\left(x+y\right)^3+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left[\left(x+y\right)^2-\left(x+y\right)+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)\)