1) Phân tích đa thức thành nhân tử:
\(\left(x+y\right)^3-x^3-y^3\)
2) Chứng minh rằng nếu:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)\(=\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2\) thì a=b=c
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a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)
\(=\left(x-y\right)\left(a+b+c\right)\)
b: \(a^m-a^{m+2}\)
\(=a^m-a^m\cdot a^2\)
\(=a^m\left(1-a^2\right)\)
\(=a^m\left(1-a\right)\left(1+a\right)\)
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
a)x2-2xy+y2+3x-3y-10
=(x2-2xy+y2)+(3x-3y)-10
=(x-y)2+3(x-y)-10
=(x-y).(x-y+3)-10
\(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a\right)+b^2c^2\left(a-c\right)+c^2a^2\left(c-a\right)\)
\(=b^2\left(a-b\right)\left(a^2-c^2\right)+c^2\left(c-a\right)\left(a^2-b^2\right)\)
\(=b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)+c^2\left(c-a\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left[-b^2\left(a+c\right)+c^2\left(a+b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(-ab^2-b^2c+ac^2+bc^2\right)\)
\(=\left(a-b\right)\left(c-a\right)\left[a\left(c^2-b^2\right)+bc\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left[a\left(c-b\right)\left(c+b\right)+bc\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)
a)\(\left(x+y\right)^3-x^3-y^3\\ =x^3+3x^2y+3xy^2+y^3-x^3-y^3\\ =3xy\left(x+y\right)\)