Đáp án:
$\begin{array}{l}
B3)\\
1)A = 6x - {x^2} - 5\\
= - \left( {{x^2} - 6x + 9} \right) + 9 - 5\\
= - {\left( {x - 3} \right)^2} + 4\\
Do:{\left( {x - 3} \right)^2} \ge 0\\
\Rightarrow - {\left( {x - 3} \right)^2} \le 0\\
\Rightarrow - {\left( {x - 3} \right)^2} + 4 \le 4\\
\Rightarrow GTLN:A = 4\,khi:x = 3\\
2)B = 4x - {x^2} + 3\\
= - \left( {{x^2} - 4x + 4} \right) + 4 + 3\\
= - {\left( {x - 2} \right)^2} + 7 \le 7\\
\Rightarrow GTLN:B = 7\,khi:x = 2\\
3)x - {x^2}\\
= - \left( {{x^2} - x} \right)\\
= - \left( {{x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4}} \right) + \dfrac{1}{4}\\
= - {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{1}{4} \le \dfrac{1}{4}\\
\Rightarrow GTLN:C = \dfrac{1}{4}\,khi:x = \dfrac{1}{2}\\
4)D = 11 - 10x - {x^2}\\
= - \left( {{x^2} + 10x + 25} \right) + 25 + 11\\
= - {\left( {x + 5} \right)^2} + 36 \le 36\\
\Rightarrow GTLN:D = 36\,khi:x = - 5\\
5)\left| {x - 4} \right|.\left( {2 - \left| {x - 4} \right|} \right)\\
= 2.\left| {x - 4} \right| - {\left( {\left| {x - 4} \right|} \right)^2}\\
= - \left( {{{\left| {x - 4} \right|}^2} - 2.\left| {x - 4} \right| + 1} \right) + 1\\
= - {\left( {\left| {x - 4} \right| - 1} \right)^2} + 1 \le 1\\
\Rightarrow GTLN:E = 1\,\\
Khi:\left| {x - 4} \right| = 1\\
\Rightarrow \left[ \begin{array}{l}
x - 4 = 1\\
x - 4 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.\\
B4)\\
a)P = 3{x^2} - 2x + 3{y^2} - 2y + 6xy - 100\\
= \left( {3{x^2} + 6xy + 3{y^2}} \right) + \left( { - 2x - 2y} \right) - 100\\
= 3\left( {{x^2} + 2xy + {y^2}} \right) - 2\left( {x + y} \right) - 100\\
= 3{\left( {x + y} \right)^2} - 2.\left( {x + y} \right) - 100\\
= {3.5^2} - 2.5 - 100\\
= - 35\\
b)Q = {x^3} + {y^3} - 2{x^2} - 2{y^2}\\
+ 3xy\left( {x + y} \right) - 4xy + 3\left( {x + y} \right) + 10\\
= \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right)\\
- 2\left( {{x^2} + 2xy + {y^2}} \right) + 3\left( {x + y} \right) + 10\\
= {\left( {x + y} \right)^3} - 2{\left( {x + y} \right)^2} + 3\left( {x + y} \right) + 10\\
= {5^3} - {2.5^2} + 3.5 + 10\\
= 125 - 50 + 15 + 10\\
= 100
\end{array}$
