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\(=b\left(9a^2+6ab+b^2\right)-2b\left(3a+b\right)\)
\(=b\left(3a+b\right)^2-2b\left(3a+b\right)\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
\(9a^2b+6ab^2+b^3-6ab-2b^2\)
\(=b\left(9a^2+6ab+b^2-6a-2b\right)\)
\(=b\left[\left(3a+b\right)^2-2\left(3a+b\right)\right]\)
\(=b\left(3a+b\right)\left(3a+b-2\right)\)
\(5a^2-5b^2-20a+20b\)
\(=5\left(a^2-b^2\right)-20\left(a-b\right)\)
\(=5\left(a-b\right)\left(a+b\right)-20\left(a-b\right)\)
\(=\left[5\left(a+b\right)-20\right]\left(a-b\right)\)
\(=\left(5a+5b-20\right)\left(a-b\right)\)
\(5a^2-5b^2-20a+20=-5.\left(b-a+2\right).\left(b+a-2\right)\)
a) Ta có: \(a^2-b^2-5a+5b\)
\(=\left(a-b\right)\left(a+b\right)-5\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-5\right)\)
b) Ta có: \(a^2-b^2-3ab^2-3a^2b\)
\(=\left(a-b\right)\left(a+b\right)-3ab\left(a+b\right)\)
\(=\left(a+b\right)\left(a-b-3ab\right)\)
a) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=2a\left(x+3\right)\left(x^2+3\right)\)
\(\left(a+b\right)^3-3ab.\left(a+b\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]=\left(a+b\right)\left(a^2+b^2-ab\right)\)
`(a+b)^3-3ab(a+b)`
`=(a+b)(a+b)^2-3ab(a+b)`
`=(a+b)[(a+b)^2-3ab]`
`=(a+b)(a^2+2ab+b^2-3ab)`
`=(a+b)(a^2-ab+b^2)`
Hửm đề sai rồi phải là:
`a^2+2b^2-3ab`
`=a^2-ab-2ab+2b^2`
`=a(a-b)-2b(a-b)`
`=(a-b)(a-2b)`
\(a^3+b^3+c^3-3ab\)
\(=a^3+ab\left(a+b\right)+b^3-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2+2ab+b^2-ab\right)-3ab\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2+b^2-ab\right)\)
= (2a-b+1)(a+2b-3)