Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+2x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+2x\right)\)
\(x^3+27x+\left(x+3\right)\left(x-9\right)\)
⇒\(x^3+27x+x^2-6x-27\)
⇒\(x^3+x^2+21x-27\)
Chịu
`#3107.101107`
`x(y - 1) + 3(y - 1)`
`= (x + 3)(y - 1)`
x(y-1)+3(y-1)
=(y-1)(x+3)
Giải thích: đặt y-1 ra làm chung .... đa thức còn x+3
a: \(3x^4-4x^3+1\)
\(=3x^4-3x^3-x^3+1\)
\(=3x^3\left(x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(3x^3-x^2-x-1\right)\)
b: \(x^3-19x-30\)
\(=x^3-4x-15x-30\)
\(=x\left(x-2\right)\left(x+2\right)-15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x-15\right)\)
\(=\left(x+2\right)\cdot\left(x-5\right)\left(x+3\right)\)
\(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
x^3-19x-30
=x^3-25x+6x-30
=x(x^2-25)+6(x-5)
=x(x+5)(x-5)+6(x-5)
=(x-5)(x^2+5x+6)
=(x-5)(x^2+2x+3x+6)
=(x-5)[x(x+2)+3(x+2)]
=(x-5)(x+2)(x+3)
\(x^3-19x-30=x^3+2x^2-2x^2-4x-15x-30\)
\(\Rightarrow x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)
\(\Rightarrow\left(x+2\right)\left(x^2-2x-15\right)\)
\(\Rightarrow\left(x+2\right)\left(x+3\right)\left(x-5\right)\)