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\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
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Ta có: \(2014S=2014\left(1+2014+2014^2+2014^3+...+2014^{2013}\right)\)
\(2014S=2014+2014^2+2014^3+2014^4+...+2014^{2014}\)
\(2014S-S=\left(2014+2014^2+2014^3+2014^4+...+2014^{2014}\right)-\left(1+2014+2014^2+2014^3+...+2014^{2013}\right)\)
\(2013S=2014^{2014}-1\)
\(S=\dfrac{2014^{2014}-1}{2013}\)
\(P-S=\dfrac{2014^{2014}}{2013}-\dfrac{2014^{2014}-1}{2013}=\dfrac{1}{2013}\)
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2014+(2014/1+2)+(2014/1+2+3)+...+(2014/1+2+3+...+2013)
=2014*(1+(1/1+2)+(1/1+2+3)+...+( 1/1+2+3+...+2013))
=2014*(1+(1/3)+(1/6)+....+(1/2027091)
=2014*2*((1/+(1/2*3)+(1/3*4).....+(1/2013*2014))
=2014*2*(1/1-1/2+1/2-1/3+1/3-1/4+.....+1/2013-1/2014)
=2014*2*(1-1/2014)
=2*(2014*2013/2014)
=2*2013
=4026
Cuối cùng cũng giải được.
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\(B=\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+....+\frac{1}{2014}\)
\(=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(1+\frac{1}{2014}\right)+1\)
\(=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2014}+\frac{1}{2015}\right)\)
\(B=\frac{2014}{1}+\frac{2013}{2}+......+\frac{1}{2014}\)
\(B=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)+1\)
\(B=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)
\(B=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
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Thầy phynit, cô @Cẩm Vân Nguyễn Thị, các bạn hok giỏi Toán: @Nguyễn Huy Tú, @Nguyễn Trần Thành Đạt, ..................
Giups em vs
Ta có :
\(2^{2014}-2^{2013}-2^{2012}-...-2-1=2^{2014}-\left(2^{2013}+2^{2012}+...+2+1\right)\)
Đặt \(A=1+2+...+2^{2012}+2^{2013}\)
\(\Leftrightarrow\)\(2A=2+2^2+...+2^{2013}+2^{2014}\)
\(\Leftrightarrow\)\(2A-A=\left(2+2^2+...+2^{2013}+2^{2014}\right)-\left(1+2+...+2^{2012}+2^{2013}\right)\)
\(\Leftrightarrow\)\(A=2^{2014}-1\)
\(\Rightarrow\)\(2^{2014}-\left(2^{2013}+2^{2012}+...+2+1\right)=2^{2014}-A=2^{2014}-\left(2^{2014}-1\right)=2^{2014}-2^{2014}+1=1\)
Vậy \(2^{2014}-2^{2013}-2^{2012}-...-2-1=1\)
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