\(f\left(x\right)=x^3-3x^2+3x-1+4=\left(x-1\right)^3+4\)

Lấy x1,x2 thuộc R sao cho x1<x2

\(A=\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{\left(x_1-1\right)^3-\left(x_2-1\right)^3}{x_1-x_2}\)

\(=\dfrac{\left(x_1-1-x_2+1\right)\left[\left(x_1-1\right)^2+\left(x_1-1\right)\left(x_2-1\right)+\left(x_2-1\right)^2\right]}{x_1-x_2}\)

\(=\left(x_1-1\right)^2+\left(x_1-1\right)\left(x_2-1\right)+\left(x_2-1\right)^2>0\)

=>A>0

Do đó: Hàm số đồng biến với x thuộc R

Do đó: \(f\left(\dfrac{2018}{2017}\right)< f\left(\dfrac{2017}{2016}\right)\)

Lời giải:

Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)

\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)

Do đó:

\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)

\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)

............

\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)

Cộng theo vế:

\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)

\(=\underbrace{1+1+1...+1}_{1008}=1008\)

Ta sẽ xét tính biến thiên của hàm số : 

Ta có \(f\left(x\right)=\left(x^3-3x^2+3x-1\right)+4=\left(x-1\right)^3+4\)

\(f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)=\left(\frac{2017}{2016}-1\right)^3-\left(\frac{2016}{2015}-1\right)^3\)

\(=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left[\left(\frac{2017}{2016}-1\right)^2+\left(\frac{2016}{2015}-1\right)^2+\left(\frac{2017}{2016}-1\right)\left(\frac{2016}{2015}-1\right)\right]\)

\(=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left(\frac{1}{2016^2}+\frac{1}{2015^2}+\frac{1}{2016}.\frac{1}{2015}\right)< 0\)

\(\Rightarrow f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)< 0\Rightarrow f\left(\frac{2017}{2016}\right)< f\left(\frac{2016}{2015}\right)\)

Ta sẽ xét tính biến thiên của hàm số : 

Ta có f\left(x\right)=\left(x^3-3x^2+3x-1\right)+4=\left(x-1\right)^3+4f(x)=(x3−3x2+3x−1)+4=(x−1)3+4

f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)=\left(\frac{2017}{2016}-1\right)^3-\left(\frac{2016}{2015}-1\right)^3f(20162017​)−f(20152016​)=(20162017​−1)3−(20152016​−1)3

=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left[\left(\frac{2017}{2016}-1\right)^2+\left(\frac{2016}{2015}-1\right)^2+\left(\frac{2017}{2016}-1\right)\left(\frac{2016}{2015}-1\right)\right]=(20161​−20151​)[(20162017​−1)2+(20152016​−1)2+(20162017​−1)(20152016​−1)]

=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left(\frac{1}{2016^2}+\frac{1}{2015^2}+\frac{1}{2016}.\frac{1}{2015}\right)&lt; 0=(20161​−20151​)(201621​+201521​+20161​.20151​)<0

\Rightarrow f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)&lt; 0\Rightarrow f\left(\frac{2017}{2016}\right)&lt; f\left(\frac{2016}{2015}\right)⇒f(20162017​)−f(20152016​)<0⇒f(20162017​)<f(20152016​)

a)\(\left(3x^2+x-2016\right)^2+4\left(x^2+506x-2017\right)^2=4\left(3x^2+x-2016\right)\cdot\left(x^2+506x-2017\right)\)

\(\Leftrightarrow\left(3x^2+x-2016\right)^2-4\left(3x^2+x-2016\right)\left(x^2+506x-2017\right)+4\left(x^2+506x-2017\right)^2=0\)

\(\Leftrightarrow\left(3x^2+x-2016-2x^2-1012x+4034\right)^2=0\)

\(\Leftrightarrow x^2-1011x+2018=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1009\end{matrix}\right.\)

Xét đa thức \(F\left(x\right)=ax^2+bx+c\)

\(F\left(0\right)=c=2016\)

\(F\left(1\right)=a+b+c=2017\Rightarrow a+b=1\)  (1)

\(F\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\)  (2)

Từ (1), (2)

\(\Rightarrow\hept{\begin{cases}a+b-a+b=-1\\a+b+a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}2b=-1\\2a=3\end{cases}}\Rightarrow\hept{\begin{cases}b=-0,5\\a=1,5\end{cases}}\)

\(\Rightarrow F\left(2\right)=1,5.2^2-0,5.2+2016=2021\)

Vậy \(F\left(2\right)=2021\).

Ta có

\(F\left(0\right)=2016\)

\(\Leftrightarrow a\cdot0^2+b\cdot0+c=2016\)

\(\Leftrightarrow0+0+c=2016\)

\(\Leftrightarrow c=2016\)

\(F\left(1\right)=2016\)

\(\Leftrightarrow a\cdot1^2+b\cdot1+c=2017\)

\(\Leftrightarrow a+b+c=2017\)

\(\Leftrightarrow a+b+2016=2017\)

\(\Leftrightarrow a+b=1\)       \(\left(1\right)\)

\(F\left(-1\right)=2018\)

\(\Leftrightarrow a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=2018\)

\(\Leftrightarrow a-b+c=2018\)

\(\Leftrightarrow a-b+2016=2018\)

\(\Leftrightarrow a-b=2\)       \(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow a=\left(1+2\right)\div2=3\div2=1.5\)

\(\Rightarrow b=1-1.5=-0.5\)

Vậy \(F\left(x\right)=1.5x^2-0.5x+2016\)

\(\Leftrightarrow F\left(2\right)=1.5\cdot2^2-0.5\cdot2+2016\)

\(=1.5\cdot4-0.5\cdot2+2016\)

\(=6-1+2016=2021\)

Vậy \(F\left(2\right)=2021\)

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