Phân tích đa thức thành nhân tử:a(a+2b)3 - b(2a+b)3
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Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
`a, 9x^2 - 16 = (3x+4)(3x-4)`
`b, 4x^2 - 12xy + 9y^2 = (2x-3y)^2`
`c, t^3-8 = (t-2)(t^2 - 2t + 4)`
`d, 2ax^3y^3 + 2a = 2a(x^3y^3 + 1) = 2a(xy+1)(x^2y^2 - xy + 1)`
a) \(\left(9x^2-16\right)=\left(3x-4\right)\left(3x+4\right)\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(t^3-8=\left(t-2\right)\left(t^2+2t+4\right)\)
d) \(2ax^3y^3+2a=2a\left(x^3y^3+1\right)\)
\(a,=ab\left(a+3\right)\\ b,=\left(x-1\right)^2\\ c,=x\left[\left(x-3\right)^2-y^2\right]=x\left(x-y-3\right)\left(x+y-3\right)\)
\(a.6x^3-9x^2=3x^2\left(2x-3\right)\\ b.20x^2y-12x^3=4x^2\left(5y-3x\right)\)
a) \(6x^3-9x^2=3x^2\left(2x-3\right)\)
b) \(20x^2y-12x^3=4x^2\left(5y-3x\right)\)
a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
Gọi a+b =x có:
a(x+b)3−b(x+a)3
=a(x3+3x2b+3xb2+b3)−b(x3+3x2a+3xa2+a3)
=ax3+3ax2b+3axb2+ab3−bx3−3bx2a−3bxa2−ba3
=(a−b)x3+(3ax2b−3bx2a)+(3axb2−3bxa2)+ab3−ba3
=(a−b)x3+3axb(b−a)+ab(b2−a2)
=−x3(b−a)+3axb(b−a)+ab(b+a)(b−a)
=−x3(b−a)+3axb(b−a)+(a2b+ab2)(b−a)
=(b−a)(−x3+3axb+a2b+ab2)
nho lik e