有几道数学题有点难,1.已知x/5=y/3,则x/(x+y)+x/(x-y)-y²/(x²-y

有几道数学题有点难,
1.已知x/5=y/3,则x/(x+y)+x/(x-y)-y²/(x²-y²)的值为________
2.（1）计算：1/(1-x)+1/(1+x)+2/(1+x²)+4/(1+x^4)
（2）猜想1/（1-x）+1/(1+x)+2/(1+x²)+4/(1+x^4)+…+1024/(1+x^1024)
3.已知abc=1,求a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)的值

1.已知x/5=y/3,则x/(x+y)+x/(x-y)-y²/(x²-y²)的值为________
x/5=y/3
x/y=5/3
x/(x+y)+x/(x-y)-y²/(x²-y²)
=x(x-y)/(x+y)(x-y)+x(x+y)/(x+y)(x-y)-y²/(x²-y²)
=[x(x-y)+x(x+y)]/(x+y)(x-y)-y²/(x²-y²)
=(x²-xy+x²+xy)/(x²-y²)-y²/(x²-y²)
=2x²/(x²-y²)-y²/(x²-y²)
=(2x²-y²)/(x²-y²)
=(2x²/y²-y²/y²)/(x²/y²-y²/y²)
=(2x²/y²-1)/(x²/y²-1)
=[2*(5/3)²-1]/[(5/3)²-1]
=(50/9-1)/(25/9-1)
=(41/9)/(16/9)
=41/16
2.（1）计算：1/(1-x)+1/(1+x)+2/(1+x²)+4/(1+x^4)
=(1+x)/(1+x)(1-x)+(1-x)/(1+x)(1-x)+2/(1+x²)+4/(1+x^4)
=[(1+x)+(1-x)]/(1+x)(1-x)+2/(1+x²)+4/(1+x^4)
=2/(1-x²)+2/(1+x²)+4/(1+x^4)
=2(1+x²)/(1-x²)(1+x²)+2(1-x²)/(1+x²)(1-x²)+4/(1+x^4)
=[2(1+x²)+2(1-x²)]/(1+x²)(1-x²)+4/(1+x^4)
=(2+2x²+2-2x²)/(1-x^4)+4/(1+x^4)
=4/(1-x^4)+4/(1+x^4)
=4(1+x^4)/(1-x^4)(1+x^4)+4(1-x^4)/(1+x^4)(1-x^4)
=[4(1+x^4)+4(1-x^4)]/(1+x^4)(1-x^4)
=(4+4x^4+4-4x^4)/(1-x^8)
=8/(1-x^8)
（2）猜想1/（1-x）+1/(1+x)+2/(1+x²)+4/(1+x^4)+…+1024/(1+x^1024)
=(1+x)/(1+x)(1-x)+(1-x)/(1+x)(1-x)+2/(1+x²)+4/(1+x^4)+…+1024/(1+x^1024)
=[(1+x)+(1-x)]/(1+x)(1-x)+2/(1+x²)+4/(1+x^4)+…+1024/(1+x^1024)
=2/(1-x²)+2/(1+x²)+4/(1+x^4)+…+1024/(1+x^1024)
=2(1+x²)/(1-x²)(1+x²)+2(1-x²)/(1+x²)(1-x²)+4/(1+x^4)+…+1024/(1+x^1024)
=[2(1+x²)+2(1-x²)]/(1+x²)(1-x²)+4/(1+x^4)+…+1024/(1+x^1024)
=(2+2x²+2-2x²)/(1-x^4)+4/(1+x^4)+…+1024/(1+x^1024)
=4/(1-x^4)+4/(1+x^4)+…+1024/(1+x^1024)
=4(1+x^4)/(1-x^4)(1+x^4)+4(1-x^4)/(1+x^4)(1-x^4)+…+1024/(1+x^1024)
=[4(1+x^4)+4(1-x^4)]/(1+x^4)(1-x^4)+…+1024/(1+x^1024)
=(4+4x^4+4-4x^4)/(1-x^8)+…+1024/(1+x^1024)
=8/(1-x^8)+…+1024/(1+x^1024)
.
=1024/(1-x^1024)+1024/(1+x^1024)
=1024(1+x^1024)/(1-x^1024)(1+x^1024)+1024(1-x^1024)/(1+x^1024)(1-x^1024)
=[1024(1+x^1024)+1024(1-x^1024)]/(1+x^1024)(1-x^1024)
=[1024+1024x^1024+1024-1024x^1024]/(1-x^2048)
=2048/(1-x^2048)
3.已知abc=1,求a/(ab+a+1)+b/(bc+b+1)+c/(ca+c+1)的值
abc=1,则a=1/bc,
则a/(ab+a+1)=1/(bc+b+1),
所以a/(ab+a+1)+b/(bc+b+1)=1/(bc+b+1)+b/(bc+b+1)
=(1+b)/(bc+b+c);
而另一个,c/(ca+c+1)可将c=1/ab代入,
则等于c/(ca+c+1)=1/(ab+a+1),
再将a=1/bc代入上式,则c/(ca+c+1)=bc/(bc+b+1),
所以,全式=1/(bc+b+1)+b/(bc+b+1)+bc/(bc+b+1)
最后=1+b+bc/bc+b+1=1.