Đặt \(f=a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-a-b\right)\)

\(=3abc+a^3+b^3+c^3-a^2b-b^2a-a^2c-b^2c-c^2a-c^2b\)

\(=a^2\left(a-b\right)+b^2\left(b-a\right)+c\left[2ab-a^2-b^2+c\left(c^2-bc-ac+ab\right)\right]\)

\(=\left(a-b\right)\left(a^2-b^2\right)-c\left(a-b\right)^2+c\left(c-a\right)\left(c-b\right)\)

\(=\left(a-b\right)^2\left(a+b+c\right)+c\left(b-c\right)\left(a-c\right)\)

\(\Rightarrow BT=\left(a-b\right)^2\left(a+b+c\right)+c\left(b-c\right)\left(a-c\right)-c\left(b-c\right)\left(a-c\right)\)

\(=\left(a+b\right)^2\left(a+b+c\right)\)

a) \(x^7+x^5+1\)

\(=x^7-x+x^5-x^2+x^2+x+1\)

\(=x\left(x^6-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x^3-1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)]

\(=\left(x^2+x+1\right)\left[x\left(x^3+1\right)\left(x-1\right)+x^2\left(x-1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left[x\left(x^4-x^3+x-1\right)+x^3-x^2+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+x^3-x^2+1\right)\)

\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)

b) \(x^5-x^4-1\)

\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)

\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)

1.

\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)

\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)

\(=\left(x^3-x^2+3x\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)

Hay đa thức trên có thể phân tích thành:

\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)

Dựa vào đó em tự tách cho phù hợp

2.

\(VT=a\left(\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+b\left(\dfrac{1}{a^2}+\dfrac{1}{c^2}\right)+c\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)

\(VT\ge\dfrac{2a}{bc}+\dfrac{2b}{ac}+\dfrac{2c}{ab}=2\dfrac{a^2+b^2+c^2}{abc}\)

\(VP=\dfrac{2\left(ab+bc+ca\right)}{abc}\)

\(\Rightarrow\dfrac{ab+bc+ca}{abc}\ge\dfrac{a^2+b^2+c^2}{abc}\)

\(\Rightarrow ab+bc+ca\ge a^2+b^2+c^2\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\le0\)

\(\Rightarrow a=b=c\)

3.

\(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\)

\(\Rightarrow\left(\dfrac{x^2-yz}{a}\right)^2=\left(\dfrac{y^2-xz}{b}\right)\left(\dfrac{z^2-xy}{c}\right)=\dfrac{\left(x^2-yz\right)^2-\left(y^2-xz\right)\left(z^2-xy\right)}{a^2-bc}\)

\(=\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}\)

Tương tự:

\(\left(\dfrac{y^2-xz}{b}\right)^2=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}\)

\(\left(\dfrac{z^2-xy}{c}\right)^2=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)

\(\Rightarrow\dfrac{x\left(x^3+y^3+z^3-3xyz\right)}{a^2-bc}=\dfrac{y\left(x^3+y^3+z^3-3xyz\right)}{b^2-ac}=\dfrac{z\left(x^3+y^3+z^3-3xyz\right)}{c^2-ab}\)

\(\Rightarrow\dfrac{x}{a^2-bc}=\dfrac{y}{b^2-ac}=\dfrac{z}{c^2-ab}\Rightarrowđpcm\)

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: \(a^m-a^{m+2}\)

\(=a^m-a^m\cdot a^2\)

\(=a^m\left(1-a^2\right)\)

\(=a^m\left(1-a\right)\left(1+a\right)\)