Bài 1:
a) \(A=x^2+3x-13\)
\(A=x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{61}{4}\)
\(A=\left(x+\frac{3}{2}\right)^2-\frac{61}{4}\ge-\frac{61}{4}\forall x\)
Dấu "=" \(\Leftrightarrow x=\frac{-3}{2}\)
b) \(B=5x^2-15x-4\)
\(B=5\left(x^2-3x-\frac{4}{5}\right)\)
\(B=5\left(x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{61}{20}\right)\)
\(B=5\left[\left(x-\frac{3}{2}\right)^2-\frac{61}{20}\right]\)
\(B=5\left(x-\frac{3}{2}\right)^2-\frac{61}{4}\ge-\frac{61}{4}\forall x\)
Dấu "=" \(\Leftrightarrow x=\frac{3}{2}\)
Bài 2:
a) \(C=2x-x^2+1\)
\(C=-\left(x^2-2x-1\right)\)
\(C=-\left(x^2-2x+1-2\right)\)
\(C=-\left[\left(x-1\right)^2-2\right]\)
\(C=2-\left(x-1\right)^2\le2\forall x\)
Dấu "=" \(\Leftrightarrow x=1\)
b) \(D=-3x^2+8x-10\)
\(D=-3\left(x^2-\frac{8}{3}x+\frac{10}{3}\right)\)
\(D=-3\left(x^2-2\cdot x\cdot\frac{4}{3}+\frac{16}{9}+\frac{14}{9}\right)\)
\(D=-3\left[\left(x-\frac{4}{3}\right)^2+\frac{14}{9}\right]\)
\(D=\frac{-14}{3}-3\left(x-\frac{4}{3}\right)^2\le\frac{-14}{3}\forall x\)
Dấu "=" \(\Leftrightarrow x=\frac{4}{3}\)