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\(\left(x+y\right)^2+\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)-3x^2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)+\left(x^2-y^2\right)-3x^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2-3x^2\)
\(=3x^2+y^2-3x^2\)
\(=y^2\)
a) (x + 3)(x2 – 3x + 9) – (54 + x3) = (x + 3)(x2 – 3x + 32 ) - (54 + x3)
= x3 + 33 - (54 + x3)
= x3 + 27 - 54 - x3
= -27
b) (2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)
= (2x + y)[(2x)2 – 2 . x . y + y2] – (2x – y)(2x)2 + 2 . x . y + y2]
= [(2x)3 + y3]- [(2x)3 - y3]
= (2x)3 + y3- (2x)3 + y3= 2y3
Bài giải:
a) (x + 3)(x2 – 3x + 9) – (54 + x3) = (x + 3)(x2 – 3x + 32 ) - (54 + x3)
= x3 + 33 - (54 + x3)
= x3 + 27 - 54 - x3
= -27
b) (2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)
= (2x + y)[(2x)2 – 2 . x . y + y2] – (2x – y)(2x)2 + 2 . x . y + y2]
= [(2x)3 + y3]- [(2x)3 - y3]
= (2x)3 + y3- (2x)3 + y3= 2y3
Bài làm
a) 2(x + y)3 + 2(x - y)3
= 2[(x + y)3 + (x - y)3]
= 2[x3 + 3x2y + 3xy2 + y3 + x3 - 3x2y + 3xy2 - y3]
= 2[(x3 + x3) + (3x2y - 3x2y) + (3xy2 + 3xy2) + (y3 - y3)]
= 2[2x3 + 6xy2]
= 4x3 + 12xy2
b)uhm... Mình sửa đề chút, thay vì là -3(x + y)2(x - y) thì mình sẽ thành +3(x + y)2(x - y)
(x - y)3 - (x + y)3 + 3(x + y)2(x - y) - 3(x + y)(x - y)2
= -[(x + y)3 - 3(x + y)2(x - y) + 3(x + y)(x - y)2 - (x - y)3]
= -[(x + y) - (x - y)]3
= -[x + y - x + y ]3
= -[y]3
= -y
\(\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(\left[\left(x+y-z\right)-\left(x+y\right)\right]^2=z^2\)
a) \(x\left(x^2-16\right)-\left(x^2+1\right)\left(x-1\right)\) =\(x^3-16x^2-x^3+x^2-x+1\)
= \(x^2-17x+1\)
b) \(\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\) = \(\left(y^4-81\right)-\left(y^4-16\right)\)
=\(-65\)
\(\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)2xy}\)
=\(\frac{x^2+5x+y^2+5y+2xy-3}{x^2+6x+y^2+6y+2xy}\)
triệt tiêu x2;y2;2xy ta được:
\(\frac{5x+5y-3}{6x+6y}=\frac{5\left(x+y\right)-3}{6\left(x+y\right)}\)
=\(\frac{5.2010-3}{6.2010}=\frac{3349}{4020}\)
`(x+y+1)^3 - (x+y-1)^3 - 6(x+y)^2`
`=(x+y+1-x-y+1)[(x+y+1)^2 + (x+y+1)(x+y-1) + (x+y-1)^2] - 6(x+y)^2`
`=2(x^2+y^2 + 2xy+2x+2y + 1 + x^2 + 2xy +y^2 - 1 + x^2 + y^2 + 1 +2xy - 2x - 2y) - 6(x^2 + 2xy + y^2)`
`=2(3x^2 + 3y^2 + 6xy +1) - 6x^2 - 12xy - 6y^2`
`=6x^2 + 6y^2 + 12xy + 2 - 6x^2 - 12xy - 6y^2`
`=2`
a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
\(=2x^2-4xy+\dfrac{15}{4}y^2\)
b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
=2x+15
a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)
b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(=2x+15\)
Nguyễn Huy Tú giúp vs
khai triển hằng đẳng thức ra rồi tính đa thức