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a: Sửa đề: x^3-x^2+5x-5
=x^2(x-1)+5(x-1)
=(x-1)(x^2+5)
b: x^3+4x^2+x-6
=x^3-x^2+5x^2-5x+6x-6
=(x-1)(x^2+5x+6)
=(x-1)(x+2)(x+3)
c: \(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
Ta có:
\(5x^2+3\left(x+y\right)^2-5y^2\)
\(\Rightarrow5x^2+3.\left(x^2+2xy+y^2\right)-5y^2\)
\(\Rightarrow5x^2+3x^2+6xy+3y^2-5y^2\)
\(\Rightarrow8x^2+6xy-2y^2\)
\(x^3+x+2=\left(x^3+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
\(b,x^4+5x^3+10x-4=\left(x^4-4\right)+\left(5x^3-10x\right)\)\(=\left(x^2+2\right)\left(x^2-2\right)+5x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2+5x\right)\)
5x^2 + 5xy - x - y
=5x.(x+y)-(x+y)
=(x+y)(5x-1)
7x - 6x^2 - 2
=-6x2+3x+4x-2
=-3x.(2x-1)+2.(2x-1)
=(2x-1)(2-3x)
$ 2x^3 - x^2 + 5x + 3 \\ = 2x^3 + x^2 - 2x^2 - x + 6x + 3 \\ = x^2(2x + 1) - x(2x + 1) + 3(2x + 1) \\ = (2x + 1)(x^2 - x + 3) $
\(2x^3-x^2+5x+3\)
= \(2x^3+x^2-2x^2-x+6x+3\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
Vì \(x^2-x+3=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+3>0\)
Nên
\(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
x^3 - x + 3x^2y + 3xy^2 + y^3 - y
=x3+y3+3x2y+3xy2-x-y
=(x+y)(x2-xy+y2)+3xy(x+y)-(x+y)
=(x+y)(x2-xy+y2+3xy-1)
=(x+y)(x2+2xy+y2-1)
=(x+y)[(x+y)2-1]
=(x+y)(x+y-1)(x+y+1)
x^2 + 5x - 6
=x2-x+6x-6
=x.(x-1)+6.(x-1)
=(x-1)(x+6)
x2 - 5x + 5
= x2 - 5(x + 1)
= x2 - \(\left[\sqrt{5\left(x+1\right)}\right]^2\)
= [x - \(\sqrt{5\left(x+1\right)}\)][x + \(\sqrt{5\left(x+1\right)}\)]