Đáp án:
$a^2+b^2+c^2+d^2-7a-15b-5c-2d\le 22$
Giải thích các bước giải:
Ta có:
$A=a^2+b^2+c^2+d^2-7a-15b-5c-2d$
$\rightarrow A+\dfrac{303}{4}=(a^2-2a.\dfrac{7}{2}+\dfrac{49}{4})+(b^2-2b.\dfrac{15}{2}+\dfrac{15^2}{4})+(c^2-2c.\dfrac{5}{2}+\dfrac{25}{4})+(d^2-2d+1)$
$\rightarrow A+\dfrac{303}{4}=(a-\dfrac{7}{2})^2+(b-\dfrac{15}{2})^2+(c-\dfrac{5}{2})^2+(d-1)^2$
Ta có:
$0\le a,b,c,d$
$3a+2b+4c+6d\le 24$
$\rightarrow \begin{cases}0\le a\le 8\\0\le b\le 12\\0\le c\le 6\\0\le d\le 4\end{cases}\rightarrow \begin{cases}-\dfrac{7}{2}\le a-\dfrac{7}{2}\le \dfrac{9}{2}\\-\dfrac{15}{2}\le b-\dfrac{15}{2}\le \dfrac{9}{2}\\-\dfrac{5}{2}\le c-\dfrac{5}{2}\le \dfrac{7}{2}\\-1\le d-1\le 3\end{cases}$
$\rightarrow \begin{cases} (a-\dfrac{7}{2})^2\le \dfrac{81}{4}\\ (b-\dfrac{15}{2})^2\le \dfrac{225}{4}\\(c-\dfrac{5}{2})^2\le \dfrac{49}{4}\\ (d-1)^2\le 9\end{cases}$
$\rightarrow(a-\dfrac{7}{2})^2+ (b-\dfrac{15}{2})^2+(c-\dfrac{5}{2})^2+(d-1)^2\le \dfrac{391}{4} $
$\rightarrow A+\dfrac{303}{4}\le \dfrac{391}{4}\rightarrow A\le 22$
