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\(A=\left(x+y+z\right)^3+\left(x-y-z\right)^3\)
\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3+x^3-3x^2\left(y+z\right)+3x\left(y+z\right)-\left(y+z\right)^3\)
\(=2x^3+6x\left(y+z\right)^2\)
\(Q=A-B+2015\)
\(=2x^3+6x\left(y+z\right)^2-2x^3-6x\left(y+z\right)^2+2015\)
=2015
\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)
\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
a) x2y3 - 1/2x4y8 = x2y3( 1 - 1/2x2y5 )
b) a2b4 + a3b - abc = ab( ab3 + a2 - c )
c) 7x( y - 4 )2 - ( y - 4 )3 = ( y - 4 )2( 7x - y + 4 )
d) -x2y2z - 6x3y - 8x4z2 - x2y2z2 = -x2( y2z + 6xy + 8x2z2 + y2z2 )
e) x3 - 4x2 + x = x( x2 - 4x + 1 )
a) \(x^3+x^2+x-3\)
\(=\left(x^3+2x^2+3x\right)-\left(x^2+2x+3\right)\)
\(=x\left(x^2+2x+3\right)-\left(x^2+2x+3\right)\)
\(=\left(x-1\right)\left(x^2+2x+3\right)\)
Bài làm:
Ta có: \(A=\left(x+y+z\right)^3+\left(x-y-z\right)^3\)
\(A=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3+x^3-3x^2\left(y+z\right)+3x\left(y+z\right)^2-\left(y+z\right)^3\)
\(A=2x^3+6x\left(y+z\right)^2=B\)
=> A = B
A = ( x + y + z )3 + ( x - y - z )3
A = [ ( x + y ) + z ]3 + [ ( x - y ) - z ]3
A = [ ( x + y )3 + 3( x + y )2z + 3( x + y )z2 + z3 ] + [ ( x - y )3 - 3( x - y )2z + 3( x - y )z2 - z3 ]
A = [ x3 + 3x2y + 3xy2 + 3x2z + 6xyz + 3y2z + 3xz2 + 3yz2 + z3 ] + [ x3 - 3x2y + 3xy2 - 3x2z + 6xyz - 3y2z + 3xz2 - 3yz2 - z3 ]
A = 2x3 + 6xy2 + 12xyz + 6xz2
B = 6x( y + z )2 + 2x3
B = 6x( y2 + 2yz + z2 ) + 2x3
B = 6xy2 + 12xyz + 6z2 + 2x3
=> A = B ( đpcm )