2) Cho a,b thỏa mãn a+2a+12=8h3–12ab+45. Tính giá trị biểu thức A=(a-2b) +2023a-4046b-22

Ta có:

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
a^{3} \ +\ 2a\ +\ 12\ =\ 8a^{3} \ -\ 12ab\ +\ 4b\\
\\
\Leftrightarrow 7a^{3} \ +\ 12ab\ -\ 2a\ -\ 4b\ -\ 12\ =\ 0\\
\\
\Leftrightarrow \ 7a^{3} \ +\ 12ab\ -\ 2( a\ +\ 2b) \ -\ 8\ =\ 0\\
\\
\Leftrightarrow \ 7a^{3} \ +\ 12ab\ -\ 2a\ -\ 4b\ +\ 4a\ -\ 8\ =\ 0\\
\\
\Leftrightarrow \ 7a^{3} \ -\ 8a\ +\ 4a\ +\ 12ab\ -\ 4b\ -\ 8\ =\ 0\\
\\
\Leftrightarrow \ \left( 7a^{3} \ -\ 8a\right) \ +\ 4( a\ +\ 3b\ -\ 2) \ =\ 0\\
\\
\Leftrightarrow \ a\left( 7a^{2} \ -\ 8\right) \ +\ 4( a\ +\ 3b\ -\ 2) \ =\ 0\\
\\
\Leftrightarrow \ a\left( 7a^{2} \ -\ 8\right) \ =\ -4( a\ +\ 3b\ -\ 2)\\
\\
\Leftrightarrow \ a\ =\ \frac{\ [ 4( 2\ -\ a\ -\ 3b)]}{7a^{2} -8}
\end{array}$

Đặt x = a - 2b, thì ta có:

$\displaystyle A\ =\ x^{2022} \ +\ \left( 2023\ +\ 2^{11}\right) x\ -\ 2^{11} .2b$

Ta cần tính x.

Ta thay a = x + 2b vào phương trình $\displaystyle 7a^{3} \ -\ 8a\ +\ 4( a\ +\ 3b\ -\ 2) \ =\ 0$, ta được:

$\displaystyle 7( x\ +\ 2b)^{3} \ -\ 8( x\ +\ 2b) \ +\ 4( x\ +\ 5b\ -\ 2) \ =\ 0$

$\displaystyle \Leftrightarrow \ 7x^{3} \ +\ 42bx^{2} \ +\ \left( 84b^{2} \ -\ 8\right) x\ +\ 16b^{3} \ -\ 20b\ +\ 8\ =\ 0$

Đặt $\displaystyle f( x) \ =\ 7x^{3} \ +\ 42bx^{2} \ +\ \left( 84b^{2} \ -\ 8\right) x\ +\ 16b^{3} \ -\ 20b\ +\ 8.$

Ta thấy $\displaystyle f( -2) \ =\ 7.( -8) \ +\ 42.b.( 4) \ +\ \left( 84b^{2} \ -\ 8\right) .( -2) \ +\ 16b^{3} \ -\ 20b\ +\ 8\ < \ 0$

và $\displaystyle f( -1) \ =\ 7.( -1) \ +\ 42.b.( 1) \ +\ \left( 84b^{2} \ -\ 8\right) .( -1) \ +\ 16b^{3} \ -\ 20b\ +\ 8\  >\ 0$

Vậy phương trình f(x) = 0 có nghiệm duy nhất trong khoảng (-2, -1).

Sử dụng định lí trung gian giá trị, ta có:

$\displaystyle -2\ < \ x\ < \ -1$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \ -2^{2022} \ < \ x^{2022} \ < \ -1^{2022}\\
\\
\Leftrightarrow \ 2^{2022} \  >\ x^{2022} \  >\ 1
\end{array}$

Vậy $\displaystyle A\ =\ x^{2022} \ +\ \left( 2023\ +\ 2^{11}\right) x\ -\ 2^{11} .2b$ nằm trong khoảng $\displaystyle \left( 2^{2022} \ +\ 2023.( -2) \ -\ 2^{11} .2,\ 2^{2022} \ +\ 2023.( -1) \ -\ 2^{1} 1.2\right)$

Vậy $\displaystyle A\ =\ 2^{2022} \ -\ 2023\ -\ 2^{12} .$