Ta có: \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right).\)

Thật vậy: \(VP=\left(x-y\right)^3+3xy\left(x-y\right)=x^3-3x^2y+3xy^2-y^3+3x^2y-3xy^2=x^3-y^3=VT\)

Áp dụng định lý trên và thay \(x-y=7;\) \(xy=15\)vào \(x^3-y^3\)ta có: \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right).\)

                                                                                                                                \(=7^3+3.15.7=343+315=658\)

\(P=x^2-5x\)

\(=x^2-5x+\frac{25}{4}-\frac{25}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{-25}{4}\forall x\)

Dấu"=" xảy ra<=> \(\left(x-\frac{5}{2}\right)^2=0\)

                   \(\Leftrightarrow x=\frac{5}{2}\)

\(P=x^2-5x\)

\(P=x^2-2x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)

\(P=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{25}{4}\)

Dấu '' = '' xảy ra 

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=0\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy Min P = 25/4 <=> x = 5/2.

a) \(A=\left(x^2-\frac{1}{2}x\right)^2+\frac{3}{4}\left(x+\frac{2}{3}\right)^2+\frac{2}{3}>0\)

Ko biết xét khoảng:v

a)CO2+CAO-->CACO3

\(\sqrt{a^2+ab+2b^2}=\sqrt{\left(\frac{3}{4}a+\frac{5}{4}b\right)^2+\frac{7}{16}\left(a-b\right)^2}\ge\sqrt{\left(\frac{3}{4}a+\frac{5}{4}b\right)^2}=\frac{3a+5b}{4}\)

Tương tự \(\sqrt{b^2+2c^2+bc}\ge\frac{3b+5c}{4};\sqrt{c^2+2a^2+ca}\ge\frac{3c+5a}{4}\)

\(\Rightarrow\sqrt{a^2+ab+2b^2}+\sqrt{b^2+2c^2+bc}+\sqrt{c^2+2a^2+ca}\ge\frac{3a+5b+3b+5c+3c+5a}{4}\)

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