Ta có a+b+c=0=>a²+b²+c²+2ab+2bc+2ca=0

=>a²+b²+c²=-2(ab+bc+ca)=>(a²+b²+c²)²=(-2ab-2bc-2ca)²

=>a⁴+b⁴+c⁴+2a²b²+2b²c²+2c²a²=4a²b²+4b²c²+4c²a²+4abc(a+b+c)=4a²b²+4b²c²+4c²a²(Do a+b+c=0)

=>a⁴+b⁴+c⁴= 2(a²b²+b²c²​+c²a²)

Ta có:

\(A=\frac{x^2-2x+206}{x^2}=1-\frac{2}{x}+\frac{206}{x^2}\)

Đặt  \(y=\frac{1}{x}\) , ta được:

\(A=1-2y+206y^2=206y^2-2y+1\)\(=\left(\sqrt{206}y\right)^2-2.\sqrt{206}.\frac{1}{\sqrt{206}}.y+\frac{1}{206}+\frac{205}{206}\)

\(A=\left(\sqrt{206}y-\frac{1}{\sqrt{206}}\right)^2+\frac{205}{206}\ge\frac{205}{206}\)  với mọi  \(y\)

Dấu  \("="\)  xảy ra  \(\Leftrightarrow\left(\sqrt{206}y-\frac{1}{\sqrt{206}}\right)^2=0\)

\(\Leftrightarrow\sqrt{206}y-\frac{1}{\sqrt{206}}=0\)

\(\Leftrightarrow\sqrt{206}y=\frac{1}{\sqrt{206}}\)

\(\Leftrightarrow y=\frac{1}{\sqrt{206}}:\sqrt{206}=\frac{1}{\sqrt{206}}.\frac{1}{\sqrt{206}}\)

\(\Leftrightarrow y=\frac{1}{206}\)

\(\Leftrightarrow x=206\)

Vậy,  \(Min\)  \(A=\frac{205}{206}\Leftrightarrow x=206\)

                                                                          ----------------------------------

a+b=17
=>(a+b)^2=289
Mà (a+b)^2=a^2+2ab+b^2; ab=72
=>a^2+b^2=289-2x72=145

(125x³ - 1) : (25x² + 5x + 1)

= (5x - 1)(25x² + 5x + 1) : (25x² + 5x + 1)

= 5x - 1