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\(\left\{{}\begin{matrix}f\left(0\right)=2014\Rightarrow c=2014\left(1\right)\\f\left(1\right)=2015\Rightarrow a+b+c=2015\left(2\right)\\f\left(-1\right)=2017\Rightarrow a-b+c=2017\left(3\right)\end{matrix}\right.\)
\(f\left(-2\right)=4a-2b+c\)
Lấy (3) nhân 3 công (2) trừ (1) nhân 2
\(f\left(-2\right)=4a-2b+c=3.2017+2015-3.2014\)
\(f\left(-2\right)=3\left(2017-2014\right)+2015=2024\)
\(y=f\left(x\right)\dfrac{2017^{2x}}{2017^{2x}+2017}\)
\(\Rightarrow\left\{{}\begin{matrix}f\left(a\right)=\dfrac{2017^{2a}}{2017^{2a}+2017}\\f\left(b\right)=\dfrac{2017^{2b}}{2017^{2b}+2017}\end{matrix}\right.\)
\(\Rightarrow f\left(a\right)+f\left(b\right)=\dfrac{2017^{2a}}{2017^{2a}+2017}+\dfrac{2017^{2b}}{2017^{2b}+2017}\)
\(=\dfrac{2017^{2a}\left(2017^{2b}+2017\right)}{\left(2017^{2a}+2017\right)\left(2017^{2b}+2017\right)}+\dfrac{2017^{2b}\left(2017^{2a}+2017\right)}{\left(2017^{2a}+2017\right)\left(2017^{2b}+2017\right)}\)
\(=\dfrac{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2\left(a+b\right)}+2017^{2b+1}}{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2b+1}+2017^2}\)
\(=\dfrac{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2\left(a+b\right)}+2017^{2b+1}}{2017^{2\left(a+b\right)}+2017^{2a+1}+2017^{2\left(a+b\right)}+2017^{2b+1}}=1\) (Vì a+b=1)
P/S:Nhìn chữ \(f\left(a\right)\) thấy khổ cho số phận mềnh quá :((
Bao giờ xong thì tag Phạm Quốc Cường lên xem lời giải nhé em
\(A=2x^2-2\ge-2\)
Dấu "=" xảy ra khi: \(x=0\)
\(B=\left|x+\dfrac{1}{3}\right|-\dfrac{1}{6}\ge-\dfrac{1}{6}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{1}{3}\)
\(C=\dfrac{\left|x\right|+2017}{2018}\ge\dfrac{2017}{2018}\)
Dấu "=" xảy ra khi: \(x=0\)
\(D=3-\left(x+1\right)^2\le3\)
Dấu "=" xảy ra khi: \(x=-1\)
\(E-\left|0,1+x\right|-1,9\le-1,9\)
Dấu "=" xảy ra khi: \(x=-0,1\)
\(F=\dfrac{1}{\left|x\right|+2017}\le\dfrac{1}{2017}\)
Dấu "=" xảy ra khi: \(x=0\)
\(f\left(1\right)=a_{2017}+a_{2016}+...+a_3+a_2+a_1+a_0\)
\(f\left(-1\right)=-a_{2017}+a_{2016}+...-a_3+a_2-a_1+a_0\)
\(f\left(1\right)+f\left(-1\right)=2\left(a_{2016}+a_{2014}+...+a_2+a_0\right)\)
\(S=\frac{f\left(1\right)+f\left(-1\right)}{2}=\frac{3^{2017}+1}{2}\)