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试证明等式(10x+y)[10x+(10一y)]=100x(x+1)+y(10一y),并利用此等式计算98x92
人气:243 ℃ 时间:2020-07-24 05:39:39
解答
1.因为(10x+y)[10x+(10一y)]
=100x²+100x-100xy+100xy+10y-y²
=100x²+100x+10y-y²
且100x(x+1)+y(10一y)
=100x²+100x+10y-y²
所以100x²+100x+10y-y²=100x²+100x+10y-y²
所以(10x+y)[10x+(10一y)]=100x(x+1)+y(10一y)
2.因为用(10x+y)[10x+(10一y)]=100x(x+1)+y(10一y)
所以令10x+y=98
10x+(10一y)=92
所以x=9
y=8
因为100x(x+1)+y(10一y)
所以原式为100*9*(9+1)+8*(10-8)
=9016
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