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1) 1
2)Ta có: 2011 x 2013 + 2012 x 2014 =8100311
20122 + 20132 - 2 =8100311 .
Vậy ta đã thấy 2 số bằng nhau
Kết luận : 2011 x 2013 + 2012 x 2014 = 20122+ 20132 - 2
1, \(B=3^{24}-\left(27^4+1\right)\left(9^6-1\right)\)
\(=\left(3^{12}\right)^2-\left(3^{12}+1\right)\left(3^{13}-1\right)\)
\(=\left(3^{12}\right)^2-\left[\left(3^{12}\right)^2-1\right]\)
\(=\left(3^{12}\right)^2-\left(3^{12}\right)^2+1\)
\(=1\)
Vậy \(B=1\)
1)
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)
\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)
\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)
\(\Leftrightarrow x=2015\)
Vậy \(S=\left\{2015\right\}\)
\(\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}\)
\(\Leftrightarrow\)\(\frac{1-x}{2013}+1=\frac{2-x}{2012}+1-\left(\frac{x}{2014}-1\right)\)
\(\Leftrightarrow\)\(\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{x-2014}{2014}\)
\(\Leftrightarrow\)\(\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}\)=0
\(\Leftrightarrow\)(2014-x)(\(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\))=0
\(\Leftrightarrow\)2014-x=0(do 1/2013 -1/2012 -1/2014)
\(\Leftrightarrow\)x=2014
\(\Leftrightarrow\frac{2-x}{2012}+1=\frac{1-x}{2013}+1+1-\frac{x}{2014}\)
\(\Leftrightarrow\frac{2014-x}{2012}=\frac{2014-x}{2013}+\frac{2014-x}{2014}\)
\(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow2014-x=0\Rightarrow x=2014\)
2) xét tử ta có
2014+2013/2+2012/3+...+2/2013+1/2014
=(1+2013/2)+(1+2012/3)+...+(1+2/2013)+(1+1/2014)+1
=2015/2+2015/3+...+2015/2013+2015/2014+2015/2015
=2015(1/2+1/3+...+1/2013+1/2014+1/2015) (1)
mà mẫu bằng 1/2+1/3+1/4+...+1/2014+1/2015 (2)
từ (1),(2)=> phân thức trên =2015
