Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)

a, Theo bài ra ta có \(\hept{\begin{cases}f\left(0\right)=c=0\\f\left(1\right)=a+b+c=2013\\f\left(-1\right)=a-b+c=2012\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=2013\\a-b=2012\end{cases}}\)

Cộng vế với vế \(a+b+a-b=2013+2012\Leftrightarrow2a=4025\Leftrightarrow a=\frac{4025}{2}\)

\(\Rightarrow b=\frac{4025}{2}-2012=\frac{1}{2}\)

Vậy \(a=\frac{4025}{2};b=\frac{1}{2};c=0\)

Ta xét : \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{3x^2-3x+1}=\frac{\left(x+1-x\right)\left(x^2+x^2-2x+1+x^2-x\right)}{3x^2-3x+1}=\frac{3x^2-3x+1}{3x^2-3x+1}=1\)

Áp dụng ta có : 

\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+\left[f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)\right]+...+\left[f\left(\frac{1006}{2012}\right)+f\left(\frac{1006}{2012}\right)\right]\)

\(=1+1+...+1\)(Có tất cả 1006 số 1)

\(=1006\)

sai rồi bạn ơi

câu a

ta có \(\hept{\begin{cases}f\left(0\right)=c=0\\f\left(1\right)=a+b+c=2013\\f\left(-1\right)=a-b+c=2012\end{cases}\Rightarrow\hept{\begin{cases}c=0\\a=2012,5\\b=0,5\end{cases}}}\)

câu b , do \(f\left(-2\right)=f\left(3\right)\Leftrightarrow4a-2b+c=9a+3b+c=2036\)

\(f\left(1\right)=a+b+c=2012\Rightarrow\hept{\begin{cases}a=4\\b=-4\\c=2012\end{cases}}\)do đó \(f\left(x\right)=4x^2-4x+2012=\left(2x-1\right)^2+2011>0\)với mọi x,

Bài 1 : làm tương tự với bài 2;3 nhé

Ta có : \(f\left(0\right)=c=2010;f\left(1\right)=a+b+c=2011\)

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

\(f\left(-1\right)=a-b+c=2012\Rightarrow f\left(-1\right)=a-b=2\)

\(\Rightarrow a+b=1;a-b=2\Rightarrow2a=3\Leftrightarrow a=\dfrac{3}{2};b=\dfrac{3}{2}-2=-\dfrac{1}{2}\)

Vậy \(f\left(-2\right)=4a-2b+c=\dfrac{4.3}{2}-2\left(-\dfrac{1}{2}\right)+2010=6+1+2010=2017\)

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

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

Ta có \(f\left(x\right)+f\left(1-x\right)=1\) khi đó

\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)

\(=1+1+...+1+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\frac{1}{2}+3.\left(\frac{1}{2}\right)^2}=1005+\frac{1}{2}=\frac{2011}{2}\)

Ta có: \(F\left(x\right)=\frac{x^3}{1-3x+3x^2}\)

\(\Leftrightarrow F\left(1-x\right)=1-\frac{x^3}{1-3x+3x^2}\)

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

\(=\frac{\left(1-x\right)^3}{1-3x+3x^2}\)

Ta có: \(F\left(x\right)+F\left(1-x\right)\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3x+3x^2}\)

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

\(\Leftrightarrow F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)=1\)

...

\(F\left(\frac{1005}{2012}\right)+F\left(\frac{1007}{2012}\right)=1\)

Do đó: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)

\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)

\(=1+1+...+F\left(\frac{1}{2}\right)\)

\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)

\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)

\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)

\(=1005+\frac{1}{2}=\frac{2011}{2}\)

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

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

Do đó:

\(A=f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)+...+f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)+f\left(\frac{1}{2}\right)\)

\(=f\left(\frac{1}{2012}\right)+f\left(1-\frac{1}{2012}\right)+...+f\left(\frac{1005}{2012}\right)+f\left(1-\frac{1005}{2012}\right)+f\left(\frac{1}{2}\right)\)

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

\(=1005+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\left(\frac{1}{2}\right)+3.\left(\frac{1}{2}\right)^2}=...\)

Đễ dàng chưng minh được

\(f\left(1-x\right)=1-f\left(x\right)\)

\(\Rightarrow f\left(1-x\right)+f\left(x\right)=1\)

\(\Rightarrow A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+\left[f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)

\(=1005+f\left(\frac{1006}{2012}\right)\)

Làm nôt