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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
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
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: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)
\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)
\(=1+1+...+F\left(\frac{1}{2}\right)\)
\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)
\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)
\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)
\(=1005+\frac{1}{2}=\frac{2011}{2}\)
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)
\(=1006\)
\(F=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{2012}-\sqrt{2011}}{\left(\sqrt{2012}+\sqrt{2011}\right)\left(\sqrt{2012}-\sqrt{2011}\right)}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2012}-\sqrt{2011}}{2012-2011}\)
\(F=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2012}-\sqrt{2011}\)
\(F=\sqrt{2012}-\sqrt{1}\)
\(F=\sqrt{2012}-1\)
Điều kiện \(\hept{\begin{cases}x-2011>0\\y-2012>0\\z-2013>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2011\\y>2012\\z>2013\end{cases}}}\)
\(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{\sqrt{x-2011}}-\frac{1}{x-2011}+\frac{1}{\sqrt{y-2012}}-\frac{1}{y-2012}+\frac{1}{\sqrt{z-2013}}-\frac{1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2011}-\frac{1}{\sqrt{x-2011}}+\frac{1}{4}\right)+\left(\frac{1}{y-2012}-\frac{1}{\sqrt{y-2012}}+\frac{1}{4}\right)+\left(\frac{1}{z-2013}-\frac{1}{\sqrt{z-2013}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2011}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{y-2012}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{z-2013}}-\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-2011}}=\frac{1}{4}\\\frac{1}{\sqrt{y-2012}}=\frac{1}{4}\\\frac{1}{\sqrt{z-2013}}=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2011=16\\y-2012=16\\z-2013=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=2027\\y=2028\\z=2029\end{cases}}}\)
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
Xét \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3+3x+3-6x+3x^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Thay vào ta tính được:
\(A=\left[f\left(\frac{1}{2020}\right)+f\left(\frac{2019}{2020}\right)\right]+...+\left[f\left(\frac{1009}{2020}\right)+f\left(\frac{1011}{2020}\right)\right]+f\left(\frac{1010}{2020}\right)\)
\(A=1+...+1+f\left(\frac{1010}{2020}\right)\) (với 1009 số 1)
\(A=1009+f\left(\frac{1}{2}\right)=1009+\frac{\left(\frac{1}{2}\right)^3}{1-3\cdot\frac{1}{2}+3\cdot\left(\frac{1}{2}\right)^2}\)
\(A=1009+\frac{1}{2}=\frac{2019}{2}\)
Vậy \(A=\frac{2019}{2}\)
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 \(\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+\left(1-x\right)^2}=\frac{\left(1-x\right)^3}{x^2-3x+1}\)