Question

Cho hàm số \(y=f\left(x\right)=\dfrac{2017^{2x}}{2017^{2x}+2017}\)

Tính \(f\left(a\right)+f\left(b\right)\) biết \(a+b=1\)

(đăng lại hộ cho thằng bạn Phạm Quốc Cường cho dễ nhìn)

Answer 1

\(y=f\left(x\right)\dfrac{2017^{2x}}{2017^{2x}+2017}\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(a\right)=\dfrac{2017^{2a}}{2017^{2a}+2017}\\f\left(b\right)=\dfrac{2017^{2b}}{2017^{2b}+2017}\end{matrix}\right.\)

\(\Rightarrow f\left(a\right)+f\left(b\right)=\dfrac{2017^{2a}}{2017^{2a}+2017}+\dfrac{2017^{2b}}{2017^{2b}+2017}\)

\(=\dfrac{2017^{2a}\left(2017^{2b}+2017\right)}{\left(2017^{2a}+2017\right)\left(2017^{2b}+2017\right)}+\dfrac{2017^{2b}\left(2017^{2a}+2017\right)}{\left(2017^{2a}+2017\right)\left(2017^{2b}+2017\right)}\)

\(=\dfrac{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2\left(a+b\right)}+2017^{2b+1}}{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2b+1}+2017^2}\)

\(=\dfrac{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2\left(a+b\right)}+2017^{2b+1}}{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2\left(a+b\right)}+2017^{2b+1}}=1\) (Vì a+b=1)

P/S:Nhìn chữ \(f\left(a\right)\) thấy khổ cho số phận mềnh quá :((

Answer 2

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