Bài 3:
a: \(M=x^2-4x+5\)
\(=x^2-4x+4+1\)
\(=\left(x-2\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi x=2
b: \(N=y^2-y-3\)
\(=y^2-2\cdot y\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{13}{4}\)
\(=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\forall y\)
Dấu '=' xảy ra khi \(y=\dfrac{1}{2}\)