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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{{{2}^{x}}}{{{2}^{x}}+2}.$ Tổng $f\left( 0 \right)+f\left( \dfrac{1}{10} \right)+f\left( \dfrac{2}{10} \right)+...+f\left( \dfrac{18}{10} \right)+f\left( \dfrac{19}{10} \right)$ bằng
A. $\dfrac{19}{2}$
B. $\dfrac{59}{6}$
C. 10
D. $\dfrac{28}{3}$
A. $\dfrac{19}{2}$
B. $\dfrac{59}{6}$
C. 10
D. $\dfrac{28}{3}$
$f\left( x \right)=\dfrac{{{2}^{x}}}{{{2}^{x}}+2}.$
$f\left( 2-x \right)=\dfrac{{{2}^{2-x}}}{{{2}^{2-x}}+2}=\dfrac{\dfrac{{{2}^{x}}}{{{2}^{x}}}}{\dfrac{{{2}^{2}}}{{{2}^{x}}}+2}=\dfrac{4}{4+{{2.2}^{x}}}=\dfrac{2}{2+{{2}^{x}}}.$
$\Rightarrow f\left( x \right)+f\left( 2-x \right)=1.$
Nên $f\left( 0 \right)+f\left( \dfrac{1}{10} \right)+f\left( \dfrac{2}{10} \right)+...+f\left( \dfrac{18}{10} \right)+f\left( \dfrac{19}{10} \right)=f\left( 0 \right)+9.1+f\left( \dfrac{10}{10} \right)=\dfrac{1}{3}+9.1+\dfrac{1}{2}=\dfrac{59}{6}.$
$f\left( 2-x \right)=\dfrac{{{2}^{2-x}}}{{{2}^{2-x}}+2}=\dfrac{\dfrac{{{2}^{x}}}{{{2}^{x}}}}{\dfrac{{{2}^{2}}}{{{2}^{x}}}+2}=\dfrac{4}{4+{{2.2}^{x}}}=\dfrac{2}{2+{{2}^{x}}}.$
$\Rightarrow f\left( x \right)+f\left( 2-x \right)=1.$
Nên $f\left( 0 \right)+f\left( \dfrac{1}{10} \right)+f\left( \dfrac{2}{10} \right)+...+f\left( \dfrac{18}{10} \right)+f\left( \dfrac{19}{10} \right)=f\left( 0 \right)+9.1+f\left( \dfrac{10}{10} \right)=\dfrac{1}{3}+9.1+\dfrac{1}{2}=\dfrac{59}{6}.$
Đáp án B.