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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{{{2025}^{x}}}{45+{{2025}^{x}}},x\in \mathbb{R}$ và hai số $a, b$ thỏa mãn $a+b=3$. Tính $f\left( a \right)+f\left( b-2 \right)$.
A. $-1$
B. $2$
C. $-2$
D. $$ $1$
A. $-1$
B. $2$
C. $-2$
D. $$ $1$
Ta có $a+(b-2)=1$.
Xét $m+n=1$.
$f\left( m \right)+f\left( n \right)=\dfrac{{{2025}^{m}}}{45+{{2025}^{m}}}+\dfrac{{{2025}^{n}}}{45+{{2025}^{n}}}=\dfrac{{{2025}^{m}}\left( 45+{{2025}^{n}} \right)+{{2025}^{n}}\left( 45+{{2025}^{m}} \right)}{\left( 45+{{2025}^{m}} \right)\left( 45+{{2025}^{n}} \right)}$
$=\dfrac{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}{45.45+45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2025}=\dfrac{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}=1$.
$f\left( a \right)+f\left( b-2 \right)=1$
Xét $m+n=1$.
$f\left( m \right)+f\left( n \right)=\dfrac{{{2025}^{m}}}{45+{{2025}^{m}}}+\dfrac{{{2025}^{n}}}{45+{{2025}^{n}}}=\dfrac{{{2025}^{m}}\left( 45+{{2025}^{n}} \right)+{{2025}^{n}}\left( 45+{{2025}^{m}} \right)}{\left( 45+{{2025}^{m}} \right)\left( 45+{{2025}^{n}} \right)}$
$=\dfrac{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}{45.45+45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2025}=\dfrac{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}{45\left( {{2025}^{m}}+{{2025}^{n}} \right)+2.2025}=1$.
$f\left( a \right)+f\left( b-2 \right)=1$
Đáp án D.