奥数题abc-ab.c=aab.c怎么解
\u6570\u5b66\u9898\u3002\u82e5abc=1,\u8bd5\u89e3\u5173\u4e8ex\u7684\u65b9\u7a0b(x/1+a+ab)+(x/1+b+bc)+(x/1+c+ac)=2006\u56e0\u4e3a ABC = 1\uff0c\u6240\u4ee5\uff1a
1 + B + BC = ABC + B + BC = B (1 + C + AC)
1 + A + AB = ABC + A + AB = A (1 + B + BC) = AB (1 + C + AC)
\u6240\u4ee5\uff1a
X/(1 + A + AB) + X/(1 + B + BC) + X/(1 + C + AC)
= X/[AB(1 + C + AC)] + X/[B(1 + C + AC)] + X/(1 + C + AC)
= [ X/(1 + C + AC) ] * (1/AB + 1/B + 1)
= [ X/(1 + C + AC) ] * [ (1 + A + AB)/AB ]
= [ X/(1 + C + AC) ] * [ AB (1 + C + AC) / AB ]
= X
= 2006
\u6240\u4ee5\uff1aX = 2006
\u5411\u91cfAB*AC=\u4e09\u5206\u4e4b\u516b\u4e09\u89d2\u5f62\u9762\u79ef
\u5373IABI*IACIcosA=(8/3)*(1/2)IABI*IACIsinA
cosA=(4/3)sinA (1)
(1)^2 1-sin^2A=(16/9)*sin^2A
\u89e3\u5f97 sinA=3/5
\u4ee3\u5165(1) cosA=4/5
1. sin(B+C)/2=cos(A/2)=\u221a[(1+cosA)/2]=3\u221a10/10
cos2A=2cos^2A-1=2*(4/5)^2-1=7/25
\u6240\u4ee5sin(B+C)/2+cos2A=3\u221a10/10+7/25
2. b=2 \u4e09\u89d2\u5f62\u9762\u79ef=(1/2)*bcsinA=3
c*(3/5)=3 c=5
\u7531\u4f59\u5f26\u5b9a\u7406 a^2=b^2+c^2-2bc*cosA=4+25-2*2*5*(4/5)=13
a=\u221a13
10*abc=1000*a+2abc
1000a=8m
125a=m
a=1,m=125
a=2,m=250
a=3,m=375
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绛旓細濡傛灉A*B*AB=BBB鍙互,鎴戜細鍋,杩欎釜棰樼瓑浜嶢AA鎴戞劅瑙夋棤棰 濡傛灉A*B*AB=BBB 鍒橝脳B脳锛10A+B锛=100B+10B+B 10A²B+AB²=111B 鈭礏鈮0,鈭翠袱杈瑰悓鏃堕櫎浠寰楋細10A²+aAB=111 A锛10A+B锛=3脳37锛堟敞鎰37鏄川鏁帮級鈭碅=3,10A+B=37 A=3,B=7 ...
