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​)

Ta có: \(\frac{1}{f\left(x\right)}-1=\frac{\left(1-x\right)^3}{x^3}\)

Xét hai số a, b dương sao cho \(a+b=1\)

Ta có: \(\hept{\begin{cases}\frac{1}{f\left(a\right)}-1=\frac{\left(1-a\right)^3}{a^3}\\\frac{1}{f\left(b\right)}-1=\frac{\left(1-b\right)^3}{b^3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{1-f\left(a\right)}{f\left(a\right)}=\frac{\left(1-a\right)^3}{a^3}\\\frac{1-f\left(b\right)}{f\left(b\right)}=\frac{a^3}{\left(1-a\right)^3}\end{cases}}\)

\(\Rightarrow\frac{1-f\left(a\right)}{f\left(a\right)}.\frac{1-f\left(b\right)}{f\left(b\right)}=1\)

\(\Rightarrow f\left(a\right)+f\left(b\right)=1\)

Áp dụng vào bài toán ta được

\(f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+...+f\left(\frac{2016}{2017}\right)\)

\(=\left[f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)\right]+\left[f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)\right]+...+\left[f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)\right]\)

\(=1+1+...+1=1008\)

Câu 2/

\(\hept{\begin{cases}2x^2-y^2+xy+3y=2\left(1\right)\\x^2-y^2=3\left(2\right)\end{cases}}\)

Ta có:

\(\left(1\right)\Leftrightarrow\left(x+y-1\right)\left(2x-y+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=1-x\\y=2x+2\end{cases}}\)

Thế ngược lại (1) giải tiếp sẽ ra nghiệm.

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 \(F\left(2\right)=2021\).