3x^2 - 6xy + 3y^2 - 12z
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`a,x^3 - 3x^2 + 1 - 3x`
`=x^3 + 1 - 3x^2 - 3x`
`=(x^3 + 1) - 3x(x+1)`
`=(x+1)(x^2 - x + 1) - 3x(x+1)`
`=(x+1)(x^2 - x + 1 - 3x)`
`=(x+1)(x^2 - 4x + 1)`
`b,x^2 + 4x - 2xy - 4y + y^2`
`=(x^2 -2xy + y^2) + (4x-4y)`
`=(x-y)^2 + 4(x-y)`
`=(x-y)(x-y+4)`
`c,3x^2 -6xy + 3y^2 - 12z^2`
`=3(x^2 -2xy +y^2 - 4z^2)`
`=3[(x-y)^2 - (2z)^2]`
`=3(x-y-2z)(x-y+2z)`
a: =x^3+1-3x^2-3x
=(x+1)(x^2-x+1)-3x(x+1)
=(x+1)(x^2-x+1-3x)
=(x+1)(x^2-4x+1)
b: =x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
c: =3(x^2-2xy+y^2-4z^2)
=3[(x-y)^2-4z^2]
=3(x-y-2z)(x-y+2z)
Ta có:
3x2-6xy +3y -12z2
=3(x2-2xy+y2-4z2)
=3[(x-y)2-4z2]
=3(x-y-2z)(x-y+2z)
Chúc bạn học tốt.
3x2-6xy+3y2-12z2
=3x2-3.2xy+3y2-3.4z2
=3(y2-2xy+y2-4z2)
=3(2y2-2xy-4z2)
= 3(x2-2xy+y2-4z2)
=3[(x-y)2-(2z)2 ]
=3(x-y-2z)(x-y+2z)
\(a,x^3-3x^2+1-3x\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x^3+1^3\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
\(c,3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left(\left(x-y\right)^2-\left(2z\right)^2\right)=3\left(x-y-2z\right)\left(x-y+2z\right)\)
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
Ta có: \(3x^2-6xy+3y^2-12z^2\)
\(=3.\left(x^2-2xy+y^2-4z^2\right)\)
\(=3.\left[\left(x-y\right)^2-4z^2\right]\)
\(=3.\left(x-y-2z\right).\left(x-y+2z\right)\)
=\(\left(\sqrt{3}x-\sqrt{3}y\right)^2-12z=\left(\sqrt{3}x-\sqrt{3}y-\sqrt{12z}\right)\left(\sqrt{3}x-\sqrt{3}y+\sqrt{12z}\right)\)