数学问题

显示全部楼层 · 2011-1-24 17:12:29

1:化简(5+1)*(5^2+1)*(5^4+1)*(5^8+1)*(5^16+1)*(5^32+1)+1/4
2,已知实数a,b,x,y满足ax+by=3和ay-bx=5,求(a^2+b^2)(x^+y^2)

千问 · 2011-1-24 17:12:29

1、可以运用平方差公式原式=(5-1)(5+1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)/(5-1) +1/4=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)/(5-1) +1/4=……=(5^64-1)/4 +1/4=(5^64)/42、(a^2+b^2)(x^2+y^2)=a^2x^2+b^2x^2+a^2y^2+b^2y^2而由已知ax+by=3
ay-bx=5将上述两式左右平方，得到
a^2x^2+b^2y^2+2abxy=9
①
a^2

千问 · 2011-1-24 17:12:29

【1】设（5+1)(52+1)(5^4+1)(5^8+1)(5^16+1)(5^32+1)=S，则4S=(5-1)(5+1)...(5^32+1)=(52-1)(52+1)...(5^32+1)=(5^4-1)(5^4+1)...(5^32+1)=(5^8-1)(5^8+1)...(5^32+1)=....=5^(64)-1.∴

千问 · 2011-1-24 17:12:29

1、(5+1)*(5^2+1)*(5^4+1)*(5^8+1)*(5^16+1)*(5^32+1)+1/4=(5-1)(5+1)*(5^2+1)*(5^4+1)*(5^8+1)*(5^16+1)*(5^32+1)/4+1/4=(5^64-1)/4+1/4=5^64/42、 ax+by=3 得 (ax+by)^2=9,即a^2x^2+b^2y^

千问 · 2011-1-24 17:12:29

1.原式=(5-1)*(5+1)*(5^2+1)*(5^4+1)*(5^8+1)*(5^16+1)*(5^32+1)/(5-1)+1/4=(5^2-1)*(5^2+1)*(5^4+1)*(5^8+1)*(5^16+1)*(5^32+1)/(5-1)+1/4=(5^4-1)*(5^4+1)*(5^8+1)*(5^16+1)*(5^32+1)/(5-1)