(2x+y)(4x^2 -2xy +y^2 )-(2x-y ) (4x^2 +2xy+y^2)
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(4x2+2xy+y2)(2x-y)-(2x-y)(4x2-2xy+y2)
=(2x3-y3)-(2x+y)(4x2-2xy+y2)
=(2x3-y3)-(2x3+y3)
=2x3-y3-2x3+y3
=0
( 2x + y ) ( 4x2 - 2xy + y2 ) - ( 2x - y ) ( 4x2 + 2xy + y2 )
= 8x3 + y3 - ( 8x3 - y3 )
= 2y3
\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=\left(8x^3-8x^3\right)+\left(y^3+y^3\right)\)
\(=2y^3\)
Sửa đề \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
a) Ta có: \(\left(y+3\right)\left(y^2-3y+9\right)-\left(60-y^3\right)\)
\(=y^3+27-60+y^3\)
\(=2y^3-33\)
b) Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
(2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)
= (2x + y)[(2x)2 – 2x.y + y2] – (2x – y)[(2x)2 + 2x.y + y2]
= [(2x)3 + y3] – [(2x)3 – y3]
= (2x)3 + y3 – (2x)3 + y3
= 2y3
a) (x+3)(x^2-3x+9)-(54+x^3)
= x^3- 3x^2+9x+3x^2-9x+27-54-x63
= -27
b) (2x + y)(4x^2 – 2xy + y^2) – (2x – y)(4x^2+ 2xy + y^2)
= (2x + y)[(2x)^2 – 2x.y + y^2] – (2x – y)[(2x)^2 + 2x.y + y^2]
= [(2x)3^3+ y^3] – [(2x)^3 – y^3]
= (2x)^3 + y^3 – (2x)^3 + y^3
= 2y^3
a)(x+3)(X^2-3x+9)-(54+x^3)
= \(x^3\)+ \(3^3 \) - 54 -\(x^3\)
= 27- 54
= -27
b)(2x+y)(4x^2-2xy+y^2)-(2x-y)(4x^2+2xy+y^2)
= \((2x)^3\) + \(y^3\) - [\((2x)^3\) - \(y^3\) ]
= \(8x^3\) + \(y^3\) - \(8x^3\) + \(y^3\)
= \(2y^3\)
(4x2 + 2xy + y2)(2x - y) - (2x + y)(4x2 - 2xy + y2) = 8x3 - y3 - 8x3 - y3 = - 2y3
\(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x-y\right)((2x)^2+2xy+y^2-\left(2x+y\right)((2x)^2-2xy+y^2\)
\(=[\left(2x\right)^3-y^3]-[\left(2x\right)^3+y^3]\)
\(=\left(2x\right)^3-y^3-\left(2x\right)^3+y^3\)
\(=-2y^3\)