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\(\frac{a}{b}=\frac{c}{d}\)=>\(\left(\frac{a}{b}\right)^{2013}=\left(\frac{c}{d}\right)^{2013}\)
=>\(\frac{a^{2013}}{b^{2013}}=\frac{c^{2013}}{d^{2013}}\)=>\(\frac{2.a^{2013}}{2.b^{2013}}=\frac{5.c^{2013}}{5.d^{2013}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\frac{2.a^{2013}}{2.b^{2013}}=\frac{5.c^{2013}}{5.d^{2013}}\)=\(\frac{2a^{2013}+5c^{2013}}{2b^{2013}+5d^{2013}}\)
=>\(\frac{a^{2013}}{b^{2013}}=\frac{c^{2013}}{d^{2013}}\)=\(\frac{2a^{2013}+5c^{2013}}{2b^{2013}+5d^{2013}}\) (\(\frac{2}{2}=1;\frac{5}{5}=1\)) (1)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
=>\(\left(\frac{a}{b}\right)^{2013}=\left(\frac{c}{d}\right)^{2013}=\left(\frac{a+b}{c+d}\right)^{2013}\)
=>\(\frac{a^{2013}}{b^{2013}}=\frac{c^{2013}}{d^{2013}}=\frac{\left(a+b\right)^{2013}}{\left(c+d\right)^{2013}}\) (2)
Từ (1) và (2)
=>\(\frac{2a^{2013}+5c^{2013}}{2b^{2013}+5d^{2013}}\)=\(\frac{\left(a+b\right)^{2013}}{\left(c+d\right)^{2013}}\)(đpcm)
\(D=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
Bài 2)
Ta có \(\frac{a}{b}< \frac{c}{d}\)
\(\Rightarrow ad< bc\)
Xét \(\frac{a}{b}< \frac{a+c}{b+d}\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow ab+ad< ab+bc\)
\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )
Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\) (1)
Xét \(\frac{a+c}{b+d}< \frac{c}{d}\)
\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow ad+cd< bc+cd\)
\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )
Vậy \(\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}{2013+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}}\)
Đặt \(B=2013+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}\)
\(=\left(2013-2013\right)\left(\frac{2013}{2}+1\right)+...+\left(\frac{1}{2014}+1\right)\)
\(=0+\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}\)
\(=2015\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
Thay B vào A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}{2015\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}\)
\(=\frac{1}{2015}\)
Vậy \(A=\frac{1}{2015}\)
Mẫu số của A \(=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)
\(=\left(1+1+...+1\right)+\left(\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\right)\)
(2012 số 1) (2011 phân số)
\(=\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1\)
\(=\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(=2013.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
=> \(A=\frac{1}{2013}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)}\)
\(\Rightarrow A=\frac{1}{2013}\)
Vậy \(A=\frac{1}{2013}\)
