我们定义一种新运算:Δ(a,b)=ab+a+b,◇(a,b)=a^2-ab+b^2,试求Δ[Δ(2,3),◇(3,2)]的值
\u5b9a\u4e49\u65b0\u8fd0\u7b97\u201c\u203b\u201d\uff0c\u89c4\u5b9a\uff1aa\u203bb=x/a+b+y/ab,\u5df2\u77e51\u203b2=8\uff0c2\u203b3=4\uff0c\u6c423\u203b4\u7684\u503c.a\u203bb=x/a+b+y/ab,\u5df2\u77e51\u203b2=8\uff0c a=1 ,b=2 x+2+y/2=8
2\u203b3=4\uff0c a=2 ,b=3 x/2+3+y/6=4 x=-6,y=24
3\u203b4 -6/3+4+24/3*4=4
2*\uff08-3/5\uff09=\uff082-\uff08-3/5\uff09\uff09/(-6/5)-1=-13/6-1=-18/6=-3
◇(3,2) = 3^2-3*2+2^2 = 7Δ(2,3) = 2*3 + 2 + 3 = 11
Δ[Δ(2,3),◇(3,2)] = Δ[11, 7] = 11*7 +11 + 7 = 95
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