\(B=3-x^2+2x\)
\(B=-\left(x^2-2x-3\right)\)
\(B=-\left(x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{13}{4}\right)\)
\(B=-\left[\left(x-\frac{1}{2}\right)^2-\frac{13}{4}\right]\)
\(B=\frac{13}{4}-\left(x-\frac{1}{2}\right)^2\)
mà \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow B\le\frac{13}{4}\forall x\)
Dấu "=" xảy ra <=> \(x-\frac{1}{2}=0\)<=> \(x=\frac{1}{2}\)
Vậy,........
\(B=3-x^2+2x\)
\(B=-\left[\left(x^2-2x+1\right)-4\right]\)
\(B=-\left[\left(x-1\right)^2-4\right]\)
\(B=4-\left(x-1\right)^2\le4\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-1\right)^2=0\)
\(\Leftrightarrow\)\(x=1\)
Vậy GTLN của \(B\) là \(4\) khi \(x=1\)
Chúc bạn học tốt ~
