http://www2.chinaedu.com/101resource004/wenjianku/200405/101ktb/lanmu/XF2S0124/XF2S0124.htm1.(x+y)/y-2x/ï¼x+yï¼+x^3/y(x^2-y^2)
2.a/(a-b)+a/(a+b)+2a^2/(a^2+b^2)+4a^4/(a^4+b^4)+8a^8/(a^8+b^8)
xyz=1
x y z
----- + ----- + ----- =?
xy+x+1 yz+z+1 zx+z+1
åå¼=x/(xy+x+xyz)+y/(yz+z+1)+yz/(xyz+yz+y)
=1/(y+1+yz)+y/(yz+z+1)+yz/(1+yz+y)
=(1+y+yz)/(1+y+yz)
=1
((x²-4/x²-4x+4)-(x-2/x+2))ä¹x-2/x
[(x^2-4)/(x^2-4x+4)-(x-2)/(x+2)]ä¹(x-2)/x
=[(x-2)(x+2)/(x-2)^2-(x-2)/(x+2)]ä¹(x-2)/x
=[(x+2)/(x-2)-(x-2)/(x+2)]ä¹(x-2)/x
=[(x+2)^2-(x-2)^2/(x-2)(x+2)]ä¹(x-2)/x
=[(x+2)^2-(x-2)^2]/[x(x+2)]
=[(x^2+4x+4-(x^2-4x+4)]/[x(x+2)]
=8x/[x(x+2)]
=8/(x+2)
å·²ç¥2X^2+X+4Y^2+4XY=-4åä¹1,æ±(X+Y)^-2çå¼.
2x^2+x+4y^2+4xy=-1/4
2x^2+x+4y^2+4xy+1/4=0
(x^2+4xy+4y^2)+(x^2+x+1/4)=0
(x+2y)^2+(x+1/2)^2=0
x+2y=0
x+1/2=0
解å¾
x=-1/2
y=1/4
å·²ç¥A+B+C=0ï¼æ±è¯ï¼ï¼1/B^2+C^2-A^2ï¼+1/C^2+A^2-B^2=1/C^2-A^2-B^2
a+b+c=0,
a=-(b+c),b=-(a+c),c=-(a+b).
1/(b^2+c^2-a^2)+1/(c^2+a^2-b^2)+1/(a^2+b^2-c^2)
=1/[b^2+c^2-(b+c)^2]+1/[c^2+a^2-(a+c)^2]+1/[a^2+b^2-(a+b)^2]
=1/(-2bc)+1/(-2ac)+1/(-2ab)
=-1/2*(a+b+c)/(abc)
=0
æ
ï¼1/B^2+C^2-A^2ï¼+1/C^2+A^2-B^2=1/C^2-A^2-B^2
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