题目内容
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分析:根据题意可知a=
(b+d+e),b=
(a+c+f),c=
(b+d+g),d=
(a+C+h),经过相加a+b+c+d=
(b+d+e)+
(a+c+f)+
(b+d+g)+
(a+C+h)得出a+b+c+d=e+f+g+h,进而解题.
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解答:解:根据题意,可得
a=
(b+d+e),
b=
(a+c+f),
c=
(b+d+g),
d=
(a+C+h),
∴a+b+c+d=
(b+d+e)+
(a+c+f)+
(b+d+g)+
(a+C+h),
3(a+b+c+d)=2(a+b+c+d)+(e+f+g+h),
则a+b+c+d=e+f+g+h,
所以(a+b+c+d)-(e+f+g+h)=0,
答:(a+b+c+d)-(e+f+g+h)的差是0.
a=
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b=
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c=
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d=
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∴a+b+c+d=
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3(a+b+c+d)=2(a+b+c+d)+(e+f+g+h),
则a+b+c+d=e+f+g+h,
所以(a+b+c+d)-(e+f+g+h)=0,
答:(a+b+c+d)-(e+f+g+h)的差是0.
点评:完成本题关健是据已知条件为突破口,得出等量关系式,从而解决问题
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