Đáp án:

\(\begin{array}{l}
a)A\left( x \right) = 10{x^3} - \dfrac{{28}}{5}{x^2} - 9\\
B\left( x \right) =  - 4{x^2} + 4x + 7\\
c) - \dfrac{{233}}{{20}}\\
c)\left[ \begin{array}{l}
x = \sqrt {\dfrac{3}{2}} \\
x =  - \sqrt {\dfrac{3}{2}} 
\end{array} \right.\\
d)Max = 9
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
P\left( x \right) = 5{x^3} - \dfrac{{24{x^2}}}{5} + 2x - 1\\
Q\left( x \right) = 5{x^3} - \dfrac{4}{5}{x^2} - 2x - 8\\
a)A\left( x \right) = P\left( x \right) + Q\left( x \right)\\
 = 5{x^3} - \dfrac{{24{x^2}}}{5} + 2x - 1 + 5{x^3} - \dfrac{4}{5}{x^2} - 2x - 8\\
 = 10{x^3} - \dfrac{{28}}{5}{x^2} - 9\\
B\left( x \right) = P\left( x \right) - Q\left( x \right)\\
 = 5{x^3} - \dfrac{{24{x^2}}}{5} + 2x - 1 - 5{x^3} + \dfrac{4}{5}{x^2} + 2x + 8\\
 =  - 4{x^2} + 4x + 7\\
c)Thay:x =  - \dfrac{1}{2}\\
 \to A = 10.{\left( { - \dfrac{1}{2}} \right)^3} - \dfrac{{28}}{5}.{\left( { - \dfrac{1}{2}} \right)^2} - 9\\
 =  - \dfrac{{233}}{{20}}\\
c)M\left( x \right) = A\left( x \right) - 10{x^3} - \dfrac{2}{5}{x^2} + 18\\
 = 10{x^3} - \dfrac{{28}}{5}{x^2} - 9 - 10{x^3} - \dfrac{2}{5}{x^2} + 18\\
 =  - 6{x^2} + 9\\
M\left( x \right) = 0\\
 \to  - 6{x^2} + 9 = 0\\
 \to 6{x^2} = 9\\
 \to {x^2} = \dfrac{3}{2}\\
 \to \left[ \begin{array}{l}
x = \sqrt {\dfrac{3}{2}} \\
x =  - \sqrt {\dfrac{3}{2}} 
\end{array} \right.\\
d)M\left( x \right) =  - 6{x^2} + 9\\
Do:6{x^2} \ge 0\forall x\\
 \to  - 6{x^2} \le 0\\
 \to  - 6{x^2} + 9 \le 9\\
 \to Max = 9\\
 \Leftrightarrow x = 0
\end{array}\)

Đáp án:

$\\$

`a,`

$\bullet$ `A (x) = P (x) + Q (x)`

`-> A (x) = (5x^3 - 4 4/5 x^2 + 2x-1) + (5x^3 - 4/5x^2 - 2x - 8)`

`-> A (x) = 5x^3  - 24/5 x^2 + 2x-1 + 5x^3 - 4/5x^2 - 2x - 8`

`-> A (x) = (5x^3 + 5x^3) + (-24/5x^2 -4/5x^2) + (2x-2x) + (-1-8)`

`-> A (x) = 10x^3 - 28/5 x^2 - 9`

$\bullet$ `B (x) = P (x) -  Q (x)`

`-> B (x) = (5x^3 - 4 4/5 x^2 + 2x-1) - (5x^3 - 4/5x^2 - 2x - 8)`

`-> B (x) = 5x^3 - 24/5 x^2+2x-1 - 5x^3 + 4/5x^2 + 2x + 8`

`-> B(x) = (5x^3 - 5x^3) + (-24/5x^2 + 4/5x^2) + (2x+2x) + (-1 + 8)`

`-> B (x) = -4x^2 + 4x + 7`

$\\$

`b,`

$\bullet$ `A (x) = 10x^3 - 28/5 x^2 - 9`

`-> A (-1/2) = 10. (-1/2)^3 - 28/5 . (-1/2)^2 -9`

`->A (-1/2) = 10 . (-1)/8 - 28/5 . 1/4 - 9`

`-> A (-1/2) = (-5)/4 - 7/5 - 9`

`-> A (-1/2) = (-53)/20 - 9`

`-> A (-1/2) = (-233)/20`

Vậy `A (-1/2) = (-233)/20`

$\\$

`c,`

$\bullet$ `M (x) = A (x) - 10x^3 - 2/5x^2 + 18`

`-> M (x) = (10x^3 - 28/5 x^2 - 9) - 10x^3 - 2/5x^2 + 18`

`-> M (x) = 10x^3 - 28/5 x^2 - 9- 10x^3 - 2/5x^2 + 18`

`-> M (x) = (10x^3 - 10x^3) + (-28/5x^2 - 2/5x^2) + (-9+18)`

`-> M (x)=-6 x^2 + 9`

Cho `M (x) = 0`

`-> -6x^2 + 9=0`

`-> -6x^2=0-9`

`-> -6x^2=-9`

`-> x^2=-9÷ (-6)`

`-> x^2=3/2`

`↔` \(\left[ \begin{array}{l}x^2=(\sqrt{\dfrac{3}{2} })^2\\x^2=(\sqrt{ \dfrac{-2}{3} })^2\end{array} \right.\) 

`↔`  \(\left[ \begin{array}{l}x=\sqrt{ \dfrac{2}{3} }\\x=\sqrt{ \dfrac{-2}{3} }\end{array} \right.\)

Vậy `x=\sqrt{2/3},x=\sqrt{ (-2)/3}` là 2 nghiệm của `M (x)`

$\\$

`d,`

$\bullet$ `M (x) = -6x^2 + 9`

Vì $x^2  \geqslant 0∀x$

$↔ -6x^2 \leqslant 0 ∀x$

$↔ -6x^2 + 9 \leqslant 9 ∀x$

`↔ M (x)` $\leqslant 9$

`↔ max M (x) = 9`

Dấu "`=`" xảy ra khi :

`↔x^2=0`

`↔x=0`

Vậy `max M (x)=9↔x=0`