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24.幻方是一种传统游戏，类比幻方，我们给出如图所示的方格，要使方格中横向、纵向及对角线方向上的实数相乘的结果都相等，求$(A+B)·D+C$的值。
答案
24.$\because$方格中横向、纵向及对角线方向上的实数相乘的结果都相等，
$\therefore A=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{5×\sqrt{2}}=2\sqrt{5},$
$B=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{10×\sqrt{10}}=1,$
$C=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{5×\sqrt{10}}=2,$
$D=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{10×\sqrt{2}}=\sqrt{5}.$
$\therefore(A+B)·D+C=(2\sqrt{5}+1)×\sqrt{5}+2=10+\sqrt{5}+2=12+\sqrt{5}.$
$\therefore(A+B)·D+C$ 的值为$12+\sqrt{5}.$
$\therefore A=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{5×\sqrt{2}}=2\sqrt{5},$
$B=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{10×\sqrt{10}}=1,$
$C=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{5×\sqrt{10}}=2,$
$D=\dfrac{\sqrt{2}×\sqrt{10}×5\sqrt{2}}{10×\sqrt{2}}=\sqrt{5}.$
$\therefore(A+B)·D+C=(2\sqrt{5}+1)×\sqrt{5}+2=10+\sqrt{5}+2=12+\sqrt{5}.$
$\therefore(A+B)·D+C$ 的值为$12+\sqrt{5}.$
