Cho hàm số $f\left( x...

Ta có \)">a+(b-2)=1m+n=1f\left( m \right)+f\left( n \right)=\dfrac{{{2025}^{m}}}{45+{{2025}^{m}}}+\dfrac{{{2025}^{n}}}{45+{{2025}^{n}}}=\dfrac{{{2025}^{m}}\left( 45+{{2025}^{n}} \right)+{{2025}^{n}}\left( 45+{{2025}^{m}} \right)}{\left( 45+{{2025}^{m}} \right)\left( 45+{{2025}^{n}} \right)}=\dfrac{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}{45.45+45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2025}=\dfrac{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}=1f\left( a \right)+f\left( b-2 \right)=1$