`a.`\(f\left(x\right)=4x^2+9x+1\)
\(f\left(x\right)=\left(4x^2+9x+\dfrac{81}{16}\right)-\dfrac{81}{16}+1\)
\(f\left(x\right)=\left(2x+\dfrac{9}{4}\right)^2-\dfrac{65}{16}\ge-\dfrac{65}{16}\)
Dấu "=" xảy ra `<=>2x+9/4=0`
`<=>x=-9/8`
Vậy \(Min_{f\left(x\right)}=-\dfrac{65}{16}\) khi `x=-9/8`
`b.`\(f\left(x\right)=-5x^2-12x+3\)
\(f\left(x\right)=-\left(5x^2+12x-3\right)\)
\(f\left(x\right)=-5\left(x^2+\dfrac{12}{5}x-\dfrac{3}{5}\right)\)
\(f\left(x\right)=-5\left(x^2+\dfrac{12}{5}x+\dfrac{36}{25}-\dfrac{36}{25}-\dfrac{3}{5}\right)\)
\(f\left(x\right)=-5\left(x+\dfrac{6}{5}\right)^2+\dfrac{51}{5}\le\dfrac{51}{5}\)
Dấu "=" xảy ra `<=>x+6/5=0`
`<=>x=-6/5`
Vậy \(Max_{f\left(x\right)}=\dfrac{51}{5}\) khi \(x=-\dfrac{6}{5}\)
