Question

Answer 1

a) Ta có: \(P=x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(Q=2x^2-6x\)

\(=2\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)

\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

c) Ta có: \(M=x^2+y^2-x+6y+10\)

\(=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{1}{2};-3\right)\)