\(x=79\Rightarrow80=x+1\)

\(P=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+\left(x+1\right)x+15\)

\(P=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+x^2+x+15\)

\(P=x+15=79+15=94\)

\(x^2+5y^2+2x-4xy-10y+14=x^2-4xy+4y^2+2x-4y+y^2-6y+9+5\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\)

Đặt \(t=x-2\), khi đó đa thức trở thành

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

\(=\left(t^2-4\right)\left(t^4+4t+16\right)+\left(t-2\right)^6\)

\(=\left(t-2\right)\left(t+2\right)\left(t^4+4t+16\right)+\left(t-2\right)^6\)

\(=\left(t-2\right)\left[t^5+4t^2+24t+2t^4+32+\left(t-2\right)^5\right]\)

Lưu ý: \(\left(a-b\right)^5=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5\)

\(=\left(t-2\right)\left(t^5+2t^4+4t^2+24t+32+\left(t^5-10t^4+40t^3-80t^2+80t-32\right)\right)\)\(=\left(t-2\right)\left(2t^5-8t^4+40t^3-76t^2+104t\right)\)

\(=\left(t-2\right)t\left(2t^4-8t^3+40t^2-76t+104\right)\)

\(A=x^2-2x+6=x^2-2x+1+5\ge5\)

Dấu "=" xảy ra khi \(x=1\)

\(B=x^2+y^2-x+5y+11=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+y^2+2.y.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}+11\)

\(B=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{5}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)

Dấu "=" xảy khi \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{5}{2}\end{matrix}\right.\)

\(C=2x^2+y^2-2xy+x-6=x^2-2xy+y^2+x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-6\)

\(C=\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge-\dfrac{25}{4}\)

dấu "=" xảy ra khi \(x=y=-\dfrac{1}{2}\)

\(D=x^2+10y^2-6xy-10y+26=x^2-6xy+9y^2+y^2-10y+25+1\)

\(D=\left(x-3y\right)^2+\left(y-5\right)^2+1\ge1\)

dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=15\\y=5\end{matrix}\right.\)