\(3x^4-5x^3-18x^2-3x+5\)

\(=\left(3x^4-6x^3-15x^2\right)+\left(x^3-2x^2-5x\right)-\left(x^2-2x-5\right)\)

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

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

\(3x^4-5x^3-18x^2-3x+5\)

\(=\left(3x^4-6x^3-15x^2\right)+\left(x^3-2x^2-5x\right)-\left(x^2-2x-5\right)\)

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

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

(2x-7)^2-2(2x-7)=0

2x-7=0 =>xx=7/3

2x-7-2=0=> x=9/2

\(\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)

\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)

\(\Rightarrow2\left(ab+bc+ca\right)\ge4\)

\(\Rightarrow-2\left(ab+bc+ca\right)\le-4\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\)(Đpcm)

Dấu = khi \(\hept{\begin{cases}\left(2-a\right)\left(2-b\right)\left(2-c\right)=0\\abc=0\\a+b+c=3\end{cases}}\)

\(\Rightarrow\left(a;b;c\right)=\left(2;1;0\right)\)và hoán vị.

a = 2 ( t/m )

b = 1 ( t/m )

c = 0 ( t/m )

vậy \(a^2+b^2+c^2\le5\)

\(A=x^3+y^3+xy=\left(x+y\right).\left(x^2-xy+y^2\right)+xy\)

     \(=x^2-xy+y^2+xy=x^2+y^2\)( do x + y = 1 )

ta có : \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}=\frac{1}{2}\)

Vậy Min A = 1/2 khi x = y = 1/2