Re: [解題] 高中數學
※ 引述《Xyla (1394障礙)》之銘言：
: 1.年級：資優班入學考題
: 2.科目：數學
: 3.章節：數
: 4.題目：
: 1.已知a,b,c,d均為實數,且滿足a+b+c+d=3, a^2+2b^2+3c^2+6d^2=5，求a範圍??
: 2.已知x,y,z均為實數,且滿足(x^2/y+z) + (y^2/z+x) + (z^2/x+y) = 0
: 求(x/y+z) + (y/z+x) + (z/x+y) = ?
[x^2/(y+z)] + [y^2/(z+x)] + [z^2/(x+y)] = 0
即 [x^2/(y+z)] + (2x+y+z) + [y^2/(z+x)]+ (x+2y+z) + [z^2/(x+y)] + (x+y+2z)
= 4(x+y+z)
即 (x+y+z)^2*{[1/(y+z)] + [1/(z+x)] + [1/(x+y)]} = 4(x+y+z)
即 (x+y+z)*{[1/(y+z)] + [1/(z+x)] + [1/(x+y)]} = 4 ---- (1) (當x+y+z≠0時)
又令 (x/y+z) + (y/z+x) + (z/x+y) = k
即 (x/y+z) + (y/z+x) + (z/x+y) + 3 = k + 3
即 (x/y+z)+1 + (y/z+x)+1 + (z/x+y)+1 = k + 3
即 (x+y+z)*{[1/(y+z)] + [1/(z+x)] + [1/(x+y)]} = k + 3 ---- (2)
由 (1) (2) 得 4 = k+3 k = 1
∴ (x/y+z) + (y/z+x) + (z/x+y) = 1
又 當 x+y+z = 0 時 (x^2/y+z) + (y^2/z+x) + (z^2/x+y) = -(x+y+z) = 0
(x/y+z) + (y/z+x) + (z/x+y) = (-1) + (-1) + (-1) = -3
: 5.想法：完全無從下手,拜託各位幫忙解一下~~感激不盡!!
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