\(=\dfrac{1993.1+1993.1994}{1992.1995+1995.1}\)

\(=\dfrac{1993\left(1+1994\right)}{1995\left(1992+1\right)}\)

\(=\dfrac{1993.1995}{1995.1993}\)

=1

=-40.(-40).7

=1600.7

=11200

Vì \(x^2\ge0\)

Mà x²\(\ne\)0 

=> Để \(x^2+\frac{1}{x^2}+3\)nho nhat => x²=1+ .x= -1;1

=> \(x^2+\frac{1}{x^2}+3\)=1+1/1+3=1+1+3=5

=> Min \(x^2+\frac{1}{x^2}+3=5\)