Giải thích bước giải :
Ta có :
$f'(x)= 2020^x-2020^{-x}=\ln \left(2020\right)\cdot \:2020^x+\ln \left(2020\right)\cdot \:2020^{-x}>0$
$\to f(x)$ đồng biến
Mà $f(x)=-f(-x)$
$\to f(a^2+b^2+ab+2)=-f(-9a-9b)=f(9a+9b)$
$\to a^2+b^2+ab+2=9a+9b$
$\to a^2+a(b-9)+b^2-9b+2=0(*)$
Đặt $\dfrac{4a+3b+1}{a+b+10}=k\to b=\dfrac{a(k-4)+(10k-1)}{3-k}$
Thay vào *
$\to \dfrac{a^2(k^2+13-7k)+a(10k^2+32-60k)+192k^2+46-311k}{(3-k)^2}=0$
$\to a^2(k^2+13-7k)+2a(5k^2+16-30k)+192k^2+46-311k=0$
$\to \Delta'=(5k^2+16-30k)^2-(k^2+13-7k)(192k^2+46-311k)\ge 0$
$\to -\left(k-2\right)\left(k-3\right)^2\left(167k-19\right)\ge 0$
$\to \left(k-2\right)\left(167k-19\right)\le 0$
$\to \dfrac{19}{167}\le k\le 2$
$\to P\le 2$
Dấu = xảy ra khi $k=2\to \Delta =0$
$\to a=-\dfrac{5k^2+16-30k}{k^2+13-7k}=-\dfrac{5\cdot 2^2+16-30\cdot 2}{2^2+13-7\cdot 2}=8$
$\to a=8\to b=3\to a^3+b^2=521\to D$
