2y3 - 6y2 + 12y - 8
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Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
\(6x^2-12xy+6y^2-6x^2\)
\(=\left(6x^2-6x^2\right)+\left(-12xy+6y^2\right)\)
\(=6y^2-12xy\)
\(=6y\left(y-2x\right)\)
10x(x-y)+6xy-6y^2
=10x(x-y)+6y(x-y)
=(x-y)(10x+6y)
=2(5x+3y)(x-y)
5 x - 2 y 3 : 5 x - 10 y = 5 x - 2 y 3 : 5 x - 2 y = x - 2 y 2
a) y3 + 12y2 + 48y + 64
= (y + 4)(y2 - 4y + 16) + 12y(y + 4)
= (y + 4)(y2 - 4y + 16 + 12y)
= (y + 4)(y2 + 8y + 16)
= (y + 4)(y + 4)2
= (y + 4)3
b) y3 - 6y2 + 12y - 8
= (y - 2)(y2 + 2y + 4) - 6y(y - 2)
= (y - 2)(y2 + 2y + 4 - 6y)
= (y - 2)(y2 - 4y + 4)
= (y - 2)(y - 2)2
= (y - 2)3
Ta có: \(2y^3-6y^2+12y-8\)
\(=2\left(y^3-3y^2+6y-4\right)\)
\(=2\left(y^3-y^2-2y^2+2y+4y-4\right)\)
\(=2\left(y-1\right)\left(y^2-2y+4\right)\)
\(2y^3-6y^2+12y-8=2y^3-4y^2-2y^2+8y+4y-8\)
\(=\left(2y^3-4y^2+8y\right)-\left(2y^2-4y+8\right)\)
\(=\left(2y^2-4y+8\right)y-\left(2y^2-4y+8\right)\)
\(=\left(y-1\right)\left(2y^2-4y+8\right)\)
\(=2\left(y-1\right)\left(y^2-2y+4\right)\)