求解三道数学题！帮帮忙！

1、由等式可得，b=3a/2 c=2a
于是带入所求的那个分式中得到(2a^2-3*3a^2+4a^2)/(a^2-3a^2-4a^2)=1/2

2.、
原方程可拆分为
(m^2+2m+1)+(n^2-6n+3*3)=0
(m+1)^2+(n-3)^2=0
因为 (m+1)^2>=0 (n-3)^2>=0
所以 m=-1, n=3
所以 m*n=-3

3,
把f(1/x)化简后，不难得到，f(x)+f(1/x)=1
原式=f(1)+1+1+1……+1
=1/2+n-1
=n-1/2

1.解：设a/2=b/3=c/4=k∴a=2k b=3k c=4k
∴原式=（8k^2-36k^2+16k^2)/(4k^2-12k^2-16k^2)=-12/(-24)=1/2
2 .解：配方：（m+1)²+（n-3)²=0∴m+1=0 n-3=0 ∴m=-1 n=3 ∴m*n=-3
3 .解：∵f(x)+f(1/x)=x^2/(1+x^2)+(1/x^2)/(1+1/x^2)=x^2/(1+x^2)+1/(1+x^2)=1
∴原式=1/2+[f(2)+f(1/2)]+[f(3)+f(1/3)]+....+f(n)+f(1/n)]
=1/2+1*(n-1)
=n-1/2

1、设a/2=b/3=c/4=t
则a=2t b=3t c=4t
(2a²-3bc+c²)/(a²-2ab-c²)
=(2*4t²-3*3t*4t+16t²)/(4t²-2*2t*3t-16t²)
=-12t²/-24t²
=1/2

2、m²+2m+n²-6n+10=0
配方：(m+1)²+(n-3)²=0
所以值域m+1=0 n-3=0
m=-1 n=3
m*n=-3

3、找规律
f(n)+f(1/n)=n²/(1+n²)+(1/n)²/(1+(1/n)²)=n²/(1+n²)+1/(1+n²)=1
所以f(1)+f(2)+f(1/2)+……+f(n)+f(1/n)
=f(1)+[f(2)+f(1/2)]+[f(3)+f(1/3)]……+[f(n)+f(1/n)]
=1/2+1+1+1……+1
=1/2+n-1
=n-1/2

1.设a/2=b/3=c/4=k,则有a=2k.b=3k,c=4k

那么有(2a^2-3bc+c^2)/(a^2-2ab-c^2)=(2*4k^2-3*3k*4k+16k^2)/(4k^2-2*2k*3k-16k^2)=(-12k^2)/(-24k^2)=1/2
2.
(m+1)^2+(n-3)^2=0
m+1=0,n-3=0
m=-1,n=3
mn=-3
3.
f(n)+f(1/n)=n^2/(1+n^2)+(1/n^2)/(1+1/n^2)=n^2/(1+n^2)+1/(n^2+1)=(n^2+1)/(n^2+1)=1
故有f(1)+f(2)+f(1/2)+....+f(n)+f(1/n)=1/2+1+1+....1(共有n-1个1)=1/2+n-1=n-1/2