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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{{{2}^{x}}}{{{2}^{x}}+2}$. Tính tổng $f\left( 0 \right)+f\left( \dfrac{1}{10} \right)+...+f\left( \dfrac{19}{10} \right)$.
A. $\dfrac{59}{6}$.
B. 10.
C. $\dfrac{19}{2}$.
D. $\dfrac{28}{3}$.
A. $\dfrac{59}{6}$.
B. 10.
C. $\dfrac{19}{2}$.
D. $\dfrac{28}{3}$.
Với $a+b=2$, ta có $f\left( a \right)+f\left( b \right)=\dfrac{{{2}^{a}}}{{{2}^{a}}+2}+\dfrac{{{2}^{b}}}{{{2}^{b}}+2}$
$=\dfrac{{{2}^{a}}{{.2}^{b}}+{{2.2}^{a}}+{{2}^{a}}{{.2}^{b}}+{{2.2}^{b}}}{\left( {{2}^{a}}+2 \right)\left( {{2}^{b}}+2 \right)}=\dfrac{{{2}^{a+b}}+{{2.2}^{a}}+{{2}^{a+b}}+{{2.2}^{b}}}{{{2}^{a+b}}+{{2.2}^{a}}+{{2.2}^{b}}+4}$
$=\dfrac{4+{{2.2}^{a}}+4+{{2.2}^{b}}}{4+{{2.2}^{a}}+{{2.2}^{b}}+4}=1$.
Do đó với $a+b=2$ thì $f\left( a \right)+f\left( b \right)=1$.
Áp dụng ta được $f\left( 0 \right)+f\left( \dfrac{1}{10} \right)+...+f\left( \dfrac{19}{10} \right)$
$=f\left( 0 \right)+\left[ f\left( \dfrac{1}{10} \right)+f\left( \dfrac{19}{10} \right) \right]+\left[ f\left( \dfrac{2}{10} \right)+f\left( \dfrac{18}{10} \right) \right]+...+\left[ f\left( \dfrac{9}{10} \right)+f\left( \dfrac{11}{10} \right) \right]+f\left( 1 \right)$
$=\dfrac{1}{3}+9.1+\dfrac{2}{4}=\dfrac{59}{6}$.
$=\dfrac{{{2}^{a}}{{.2}^{b}}+{{2.2}^{a}}+{{2}^{a}}{{.2}^{b}}+{{2.2}^{b}}}{\left( {{2}^{a}}+2 \right)\left( {{2}^{b}}+2 \right)}=\dfrac{{{2}^{a+b}}+{{2.2}^{a}}+{{2}^{a+b}}+{{2.2}^{b}}}{{{2}^{a+b}}+{{2.2}^{a}}+{{2.2}^{b}}+4}$
$=\dfrac{4+{{2.2}^{a}}+4+{{2.2}^{b}}}{4+{{2.2}^{a}}+{{2.2}^{b}}+4}=1$.
Do đó với $a+b=2$ thì $f\left( a \right)+f\left( b \right)=1$.
Áp dụng ta được $f\left( 0 \right)+f\left( \dfrac{1}{10} \right)+...+f\left( \dfrac{19}{10} \right)$
$=f\left( 0 \right)+\left[ f\left( \dfrac{1}{10} \right)+f\left( \dfrac{19}{10} \right) \right]+\left[ f\left( \dfrac{2}{10} \right)+f\left( \dfrac{18}{10} \right) \right]+...+\left[ f\left( \dfrac{9}{10} \right)+f\left( \dfrac{11}{10} \right) \right]+f\left( 1 \right)$
$=\dfrac{1}{3}+9.1+\dfrac{2}{4}=\dfrac{59}{6}$.
Đáp án A.