\(B=1+3x-x^2=-\left(x^2-3x-1\right)\)
\(=-\left[x^2-2.\frac{3}{2}.x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-1\right]\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{3}{14}\le\frac{3}{14}\)
Vậy \(B_{Max}=\frac{13}{4}\)
Khi : \(x-\frac{3}{2}=0\)
\(x=\frac{3}{2}\)
\(B=1+3x-x^2\)
\(=-x^2+3x+1\)
\(=-\left(x^2-3x-1\right)\)
\(=-\left\{\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]-\left(\frac{3}{2}\right)^2-1\right\}\)
\(=-\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}-\frac{4}{4}\right]\)
\(=-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\)
\(Có:\left(x-\frac{3}{2}\right)^2\ge0\) \(\text{với mọi x}\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2\le0\text{ với mọi x}\)
\(\Rightarrow-\left(x-\frac{3}{2}\right)^2+\frac{13}{4}\le0+\frac{13}{4}=\frac{13}{4}\text{ }\text{ với mọi x}\)
\(\text{GTLN của biểu thức B là }\frac{13}{4}\)
khi \(x-\frac{3}{2}=0\) hay \(x=\frac{3}{2}\)
