Tính: 8x^3-(2x+y)*(4x^2-2xy+y^2)
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bai2 :cmr
a, a^3+b^3=(a+b)^3-3ab.(a+b)
VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)
=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)
=VT
b.a^3-b^3=(a-b)^3+3ab,(a-b)
\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)
=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)
=VT
=> ĐPCM
bài 1.
a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)
= 8x3+4x2y+2xy2-4x2y-2xy2-y3 - 8x3+4x2y-2xy2-4x2y+2xy2-y3
=-8x2y-6y3
b) = 27x3-18x2y+12xy2+18x2y-12xy2+8y3-27x3
=8y
\(Bài1:\\ a,\left(4x-1\right)\left(2x^2-x-1\right)=4x\left(2x^2-x-1\right)-\left(2x^2-x-1\right)=8x^3-4x^2-4x-2x^2+x+1=8x^3-6x^2-3x+1\\ b,\left(4x^3+8x^2-2x\right):2x\\ =2x\left(2x^2+4x-1\right):2x\\ =2x^2+4x-1\)
\(Bài2:\\ a,2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\\ b,2xy+2x+yz+z=2x\left(y+1\right)+z\left(y+1\right)=\left(y+1\right)\left(2x+z\right)\\ c,x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)
a , x(3-x)-y(y-2x)+y(y-2)-x(2y-x)
=3x-x2-y2+2xy+y2-2y-2xy+x2
=3x-(-x2+x2)+(2xy-2xy)-2y
=3x-2y
b,(x2-7)(x+2)-(2x-1)(x-14)+x(x2-2x-22)+35
=x3+2x2-7x-14-2x2+29x-14+x3-2x2-22x+35
=(x3+x3)+(2x2-2x2-2x2)-(-7x+29x-22x)-14-14+35
=2x3-2x2+7
c,(2x+y)(4x2-2xy+y2)-8x3-y3
=2xy2-4x2y+8x3+y3-2xy2+4x2y-8x3-y3
=(2xy2-2xy2)-(-4x2y+4x2y)+(8x3-8x3)+(y3-y3)
=0
Bài 1) A=(8x3+27x3):2x+3y
=[(2x)3+(3y)3]:2x+3y
=(2x)2+(3y)2
=4x2+9y2
B=(x3-27):(x-3)
=(x3-33):(x-3)
=x2-32
=x2-9
8x3-(2x+y).(4x2-2xy+y2)
=\(\left(2x\right)^3-\left(2x+y\right).\left[\left(2x\right)^2-2x.y+y^2\right]\)
= \(\left(2x\right)^3-\left[\left(2x\right)^3+y^3\right]\)
= \(\left(2x\right)^3-\left(2x\right)^3-y^3\)
= -y3
Học tốt !
\(8x^3-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-8x^3-y^3\)
\(=-y^3\)