b: Ta có: \(x^2-x+5\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{19}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\forall x\)

\(\Leftrightarrow\dfrac{2022}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}}\le\dfrac{8088}{19}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

Lời giải:

Ta có:

$x^2-3x+11=(x-\frac{3}{2})^2+\frac{35}{4}\geq \frac{35]{4}$

$\Rightarrow \frac{31}{x^2-3x+11}\leq 31:\frac{35}{4}=\frac{124}{35}$

$\Rightarrow \frac{31}{x^2-3x+11}+15\leq \frac{649}{35}$

Vậy gtln của biểu thức là $\frac{649}{35}$ khi $x=\frac{3}{2}$

a: \(x^2-2x+3\)

\(=x^2-2x+1+2=\left(x-1\right)^2+2\ge2\forall x\)

=>\(A=\frac{37}{x^2-2x+3}\le\frac{37}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(x^2-5x+10\)

\(=x^2-5x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac52\right)^2+\frac{15}{4}\ge\frac{15}{4}\forall x\)

=>\(\frac{26}{x^2-5x+10}\le26:\frac{15}{4}=26\cdot\frac{4}{15}=\frac{104}{15}\forall x\)

=>\(B=-\frac{26}{x^2-5x+10}\ge-\frac{104}{15}\forall x\)

Dấu '=' xảy ra khi \(x-\frac52=0\)

=>\(x=\frac52\)

c: \(x^2-x+6\)

\(=x^2-x+\frac14+\frac{23}{4}\)

\(=\left(x-\frac12\right)^2+\frac{23}{4}\ge\frac{23}{4}\forall x\)

=>\(\frac{2023}{x^2-x+6}\le2023:\frac{23}{4}=2023\cdot\frac{4}{23}=\frac{8092}{23}\forall x\)

=>\(C=-\frac{2023}{x^2-x+6}\ge-\frac{8092}{23}\forall x\)

Dấu '=' xảy ra khi \(x-\frac12=0\)

=>\(x=\frac12\)

d: \(x^2+x+5\)

\(=x^2+x+\frac14+\frac{19}{4}\)

\(=\left(x+\frac12\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

=>\(D=\frac{0.75}{x^2+x+5}\le\frac34:\frac{19}{4}=\frac{3}{19}\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

e: \(2x^2-x+37=2\left(x^2-\frac12x+\frac{37}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}+\frac{295}{16}\right)=2\left(x-\frac14\right)^2+\frac{295}{8}\ge\frac{295}{8}\forall x\)

=>\(\frac{13}{2x^2-x+37}\le13:\frac{295}{8}=\frac{104}{295}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

f: \(3x^2-x+19\)

\(=3\left(x^2-\frac13x+\frac{19}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}+\frac{227}{36}\right)=3\left(x-\frac16\right)^2+\frac{227}{12}\ge\frac{227}{12}\forall x\)

=>\(\frac{61}{3x^2-x+19}\le61:\frac{227}{12}=61\cdot\frac{12}{227}=\frac{732}{227}\forall x\)

=>\(-\frac{61}{3x^2-x+19}\ge-\frac{732}{227}\forall x\)

Dấu '=' xảy ra khi x-1/6=0

=>x=1/6

$\textbf{a)}$

$S=\dfrac3{2x^2+2x+3}$

$=\dfrac3{2\left(x+\dfrac12\right)^2+\dfrac52}.$

Vì $2\left(x+\dfrac12\right)^2+\dfrac52\ge\dfrac52$

$\Rightarrow S\le\dfrac3{5/2}=\dfrac65.$

Dấu ``='' khi $x=-\dfrac12.$

Vậy $\max S=\dfrac65.$

$\textbf{b)}$

$T=\dfrac5{3x^2+4x+15}$

$=\dfrac5{3\left(x+\dfrac23\right)^2+\dfrac{41}3}.$

Vì $3\left(x+\dfrac23\right)^2+\dfrac{41}3\ge\dfrac{41}3$

$\Rightarrow T\le\dfrac5{41/3}=\dfrac{15}{41}.$

Dấu ``='' khi $x=-\dfrac23.$

Vậy $\max T=\dfrac{15}{41}.$