Question

Cho hàm số y=f(x)=2017^2x/2017^2x+2017. Tính f(a)+f(b)biết a+b=1

Answer 1

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Answer 2

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Answer 3

Ta có: f(a)+f(b)=\(\dfrac{2017^{2a}}{2017^{2a}+2017}\)+\(\dfrac{2017^{2b}}{2017^{2b}+2017}\)

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

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

Do a+b=1 nên 2(a+b)=2

=> f(a)+f(b)=\(\dfrac{2017^2+2017^{2a+1}+2017^2+2017^{2b+1}}{2017^2+2017^{2a+1}+2017^2+2017^{2b+1}}\)

=1

Vậy f(a)+f(b)=1