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x3 - 19x - 30
= x3 - 4x - 15x - 30
= x(x2 - 4) - 15(x + 2)
= x(x - 2)(x + 2) - 15(x + 2)
= (x2 - 2x) (x + 2) - 15(x + 2)
= (x + 2)(x2 - 2x - 15)
= (x + 2)(x2 - 5x + 3x - 15)
= (x + 2)(x - 5)(x + 3)
a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\) *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\) cho nhanh*
b) \(a^3-b^3+c^3+3abc=\left(a^3-3a^2b+3ab^2-b^2\right)+3a^2b-3ab^2+c^3+3abc\)
\(=\left(a-b\right)^3+c^3+\left(3a^2b-3ab^2+3abc\right)\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)
c) \(a^3-b^3-c^3-3abc=\left[a^3-3a^2b+3ab^2-b^3\right]+3a^2b-3ab^2-c^3-3abc\)
\(=\left[\left(a-b\right)^3-c^3\right]+3ab\left(a-b-c\right)=\left(a-b-c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b-c\right)\)
\(=\left(a-b-c\right)\left[a^2-2ab+b^2+ac-bc+c^2+3ab\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
a,(x+1)2
b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}
c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}
\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(a^2-b^2\right)-\left(b+c\right)\left(c^2-a^2\right)+\left(a+c\right)\left(c^2-a^2\right)\)
\(=\left(a^2-b^2\right)\left(a+b-b-c\right)-\left(c^2-a^2\right)\left(b+c-c-a\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)-\left(c-a\right)\left(c+a\right)\left(b-a\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(a+b-c-a\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
1)
a) (x+y)3-(x+y)= (x+y)(x+y-1)
b) xem lại đề câu B nha bạn
2)
a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc=0
(a+b)3+c3-3ab(a+b+c)=0
(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)=0
(a+b+c)(a2+b2+c2-xy-yz-xz)=0
Suy ra: a3+b3+c3=3abc
1. a) = (x+y)3 -(x+y) =(x+y)((x+y)2 -1)
= (x+y)(x+y+1)(x+y-1)
b) = 5(( x-y)2 - 4z2)
= 5( x-y +2z)(x-y-2z)
2. áp dụng ( a+b+c)3 = .....rồi biến đổi
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)