Question

Cho \(f(x)=\frac{x^3}{1-3x+3x^2}\)Hãy tính giá trị của biểu thức sau

\(A=f(\frac{1}{2012})+f(\frac{2}{2012})+...+f(\frac{2010}{2012})+f(\frac{2011}{2012})\)

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

Answer 3

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

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

Do đó:

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

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

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

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