\(P=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Đẳng thức xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
\(Q=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
\(M=x^2+y^2-x+6y+10\)
\(=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)
P=\(\left(X-1\right)^2+4\) \(\ge4\)=> giá trị nhỏ nhất là 4
Dấu = xảy ra khi x=1
M=\(\left(X^2-X\right)+\left(y^2+6y+9\right)+1=X\left(X-1\right)+\left(Y+3\right)^2+1\ge1\)
Dấu = xảy ra khi X=1 và Y=-3