MATH: Find the function \(f(x)\) satisfying: \[ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy} \]

Find the function \(f(x)\) satisfying:

\[ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy} \]

Solution:

Put \(y=1\), and let \(f(1)=a\).

\[ af(x)-f(x)-90=\frac{10(x+1)}{x} \]

\[ (a-1)f(x)=100+\frac{10}{x} \]

Therefore,

\[ f(x)=A+\frac{B}{x} \]

Substitute into the original equation:

\[ \left(A+\frac{B}{x}\right) \left(A+\frac{B}{y}\right) -\left(A+\frac{B}{xy}\right) =90+\frac{10}{x}+\frac{10}{y} \]

Comparing coefficients:

\[ A^2-A=90 \]

\[ AB=10 \]

\[ B^2-B=0 \]

From \(B(B-1)=0\), we get \(B=1\) because \(B=0\) cannot satisfy \(AB=10\).

Hence:

\[ A=10 \]

Therefore:

\[ \boxed{f(x)=10+\frac{1}{x}} \]

Verification:

\[ \left(10+\frac1x\right)\left(10+\frac1y\right) -\left(10+\frac1{xy}\right) \]

\[ =90+\frac{10}{x}+\frac{10}{y} \]

\[ =90+\frac{10(x+y)}{xy} \]

Therefore, the function is proven:

\[ \boxed{f(x)=10+\frac1x} \]

This problem, I picked from Vietnamese book which related to finding function. It was a problem of Turkey National Olympiad First Round 2008 which also a difficult problem in Mathematics Calculation.
In Vietnamese language this problem:

MATH