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题目没错的话，不是很简单。
如果是，a+b+c=0，1/(a+1)+1/(b+2)+1/(c+3)=0的话，简单：
1/(a+1)+1/(b+2)+1/(c+3)=0 => [(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)]/[(a+1)(b+2)(c+3)]=0
=>(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0
(a+1)^2+(b+2)^2+(c+3)^2
=[(a+1)+(b+2)+(c+3)]^2-2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]
=(a+b+c+6)^2
=36
过程？？
1\/(a+1)+1\/(b+2)+1\/(c+3)=0,则2a+2b+2c+12等于多少,如何求解
a+b+c=0,所以(a+1)+(b+2)+(c+3)=6,1\/a+1+1\/b+2+1\/c+3=0,得(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)=0 那么，（a+1)的平方+（b+2)的平方+（c+3)的平方 ＝〔(a+1)+(b+2)+(c+3)〕＾2－2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕＝36－0 ...
如果a+b+c=0,a+1分之1+b+1分之1+c+3分之1=0,求(a+1)平方+(b+2)平方+...
(a+1)^2+(b+2)^2+(c+3)^2+2(a+1)(b+2)+2(a+1)(c+3)+2(b+2)(c+3)=36 (a+1)^2+(b+2)^2+(c+3)^2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36...1 1\/(a+1)+1\/（b+2）+1\/(c+3)=0 去分母 (a+1)(b+2)+(a+1)(c+3)+(b+2)(c...
. 如果a+b+c=0,1\/(a+1)+1\/(b+2)+1\/(c+3)=0,那么(a+1)^2+(
记x=a+1,y=b+2,z=c+3 得x+y+z=6 1\/x+1\/y+1\/z=0 即xy+yz+zx=0 所以所求为 x^2+y^2+z^2 =(x+y+z)^2-2(xy+yz+zx)=36
若a+b+c=0,1\/(a+1)+1\/(b+2)+1\/(c+3)=0,则(a+1)+(b+2)(c+3)的值为?
又∵a+b+c=0 ∴b+c=-a ∴1\/(a+1)=-(5-a)\/(b+2)(c+3)∴(a+1)(5-a)=-(b+2)(c+3)∴(a+1)+(b+2)(c+3)=a+1-(a+1)(5-a)=(a+1)(1-5+a)=(a+1)(a-4)
已知a+b+c=1 且 1\/(a+2)+1\/(b+3)+1\/(c+4)=0,求(a+2)^+(b+3)^+(c+4...
设x=a+2,y=b+3,z=c+4 则x+y+z=10 1\/x +1\/y +1\/z =0，即xy+yz+xz=0 所以x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=100 即(a+2)^2+(b+3)^2+(c+4)^2=100
高中数学题求解:已知a+b+c=1,求1\/3a+2 + 1\/3b+2 + 1\/3c+2 的最小值...
解法一:a、b、c为正实数,且a+b+c=1 故由柯西不等式得 [(3a+2)+(3b+2)+(3c+2)]*[1\/(3a+2)+1\/(3b+2)+1\/(3c+2)]>=(1+1+1)^2 --->[3(a+b+c)+6]*[1\/(3a+2)+1\/(3b+2)+1\/(3c+2)]>=9 --->[3×1+6]*[1\/(3a+2)+1\/(3b+2)+1\/(3c+2)]>=9...
a+b+c=0,求证1\/(b^2+c^2-a^2)+1\/(c^2+a^2-b^2)+1\/(a^2+b^2-c^2)=0
解：因为a+b+c=0 所以a=-b-c b=-a-c c=-a-b a^2=(b+c)^2=b^2+2bc+c^2 b^2=(a+c)^2=a^2+2ac+c^2 c^2=(a+b)^2=a^2+2ab+b^2 代入就可以得到：1\/(b^2+c^2-a^2)+1\/(c^2+a^2-b^2)+1\/(a^2+b^2-c^2)=0 =1\/(b^2+c^2-b^2-2bc-c^2)...
已知A+B+C=0,求证a(1\/b+1\/c)+B(1\/a+1\/c)+c(1\/a+\/b)+3=0
a(1\/b+1\/c)=a(c+b)\/bc b(1\/a+1\/c)=b(a+c)\/ac c(1\/a+1\/b)=c(a+b)\/ab 三式相加 原式=a^2(c+b)+b^2(a+c)+c^2(a+b)\/abc 而因为a+b+c=0,(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac 所以a=(-b-c) b=(-a-c) c=(-a-b) 所以原式=-(b+c)^2-...
已知a,b,c≥0,且a+b+c=1,求证1\/a^2+a+1+1\/b^2+b+1+1\/c^2+c+1≥7\/3
1\/a+1\/b+1\/c)+c(1\/a+1\/b+1\/c)=0 (a+b+c)*(1\/a+1\/b+1\/c)=0 a+b+c=0 或1\/a+1\/b+1\/c=0 (bc+ac+ab)\/(abc)=0 ab+ac+bc=0 a^2+b^2+c^2=1 a^2+b^2+c^2+2ab+2ac+2bc=1+0 (a+b+c)^2=1 a+b+c=1或-1 综上所述a+b+c=0或1或-1 ...
若a+b+c=0,且abc≠0,求a(1\/b+1\/c)+b(1\/a+1\/c)+c(1\/a+1\/b)+2的值
因a+b+c=0，所以a+b=-c;b+c=-a;a+c=-b;则原式=a\/b+a\/c+b\/a+b\/c+c\/a+c\/b+2 =(a+c)\/b+(a+b)\/c+(b+c)\/a+2 =(-b)\/b+(-c)\/c+(-a)\/a+2 =-1+(-1)+(-1)+2 =-1
