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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)
\(=1006\)
Ta thấy: \(f\left(x\right)=\frac{x^3}{1-3x+x^2}\)
\(f\left(1-x\right)=\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+\left(1-x\right)^2}\)\(=\frac{\left(1-x\right)^3}{x^2-3x+1}\)
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3+\left(1-x\right)^3}{x^2-3x+1}\)=1
Do đó: \(f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)=1\)
\(f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)=1\)
....
\(f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)=1\)
=>A=1+1+1+...+1+\(f\left(\frac{1006}{2012}\right)\)=\(\frac{2009}{2}\)
(1005 số 1)
bn ơi cho mình hỏi dòng thứ 2 á tại sao \(\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+\left(1-x\right)^2}=\frac{\left(1-x\right)^3}{x^2-3x+1}\)
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
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=2012\)
\(\Leftrightarrow\)\(\frac{x-1-2012}{2012}+\frac{x-2-2011}{2011}+\frac{x-3-2010}{2010}+...+\frac{x-2012-1}{1}=0\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\ne0\)
Nên \(x-2013=0\)
\(\Leftrightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt ~
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1+2012=2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
\(\Leftrightarrow x=2013\)
\(F=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{2012}-\sqrt{2011}}{\left(\sqrt{2012}+\sqrt{2011}\right)\left(\sqrt{2012}-\sqrt{2011}\right)}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2012}-\sqrt{2011}}{2012-2011}\)
\(F=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2012}-\sqrt{2011}\)
\(F=\sqrt{2012}-\sqrt{1}\)
\(F=\sqrt{2012}-1\)
Áp dụng BĐT Cô - si ngược dấu :
\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4\left(x-2010\right)}\le\frac{4+\left(x-2010\right)}{4}\)
\(\Rightarrow\sqrt{x-2010}-1\le\frac{4+\left(x-2010\right)}{4}-1=\frac{x-2010}{4}\)
\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}\le\frac{1}{4}\)
Hoàn toàn tương tự với những phân thức còn lại
\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2010=4\\x-2011=4\\z-2012=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=2014\\y=2015\\z=2016\end{cases}}}\)
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
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: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)
\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)
\(=1+1+...+F\left(\frac{1}{2}\right)\)
\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)
\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)
\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)
\(=1005+\frac{1}{2}=\frac{2011}{2}\)