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Ta có :
\(\frac{a+2015}{a-2015}=\frac{b+2016}{b-2016}\Rightarrow\frac{a+2015}{b+2016}=\frac{a-2015}{b-2016}=\frac{a+2015+a-2015}{b+2016+b-2016}=\frac{2a}{2b}=\frac{a}{b}\)
Bạn xét tích chéo là ra
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
Ta có \(B=\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)
\(\Rightarrow B=1+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)\)
\(\Rightarrow B=\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}\)
\(\Rightarrow B=2017.\left(\frac{1}{2017}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
\(\Rightarrow B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)= 2017
~ Chúc bạn học tốt
Có: \(\sqrt{2015}< \sqrt{2016}\)
=>\(\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2016}}\)
=>\(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>0\)
=>\(\sqrt{2015}+\sqrt{2016}+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
=>\(\left(\sqrt{2015}+\frac{1}{\sqrt{2015}}\right)+\left(\sqrt{2016}-\frac{1}{\sqrt{2016}}\right)>\sqrt{2015}+\sqrt{2016}\)
=>\(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
ta có : \(\frac{a+2016}{a-2016}=\frac{b+2015}{b-2015}\)
=> \(\frac{a+2016}{b+2015}=\frac{a-2016}{b-2015}\)
=> Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a+2016}{b+2015}=\frac{a-2016}{b-2015}=\frac{a+2016+a-2016}{b+2015+b-2015}=\frac{2a}{2b}=\frac{a}{b}\)
=> \(\frac{a+2016}{b+2015}=\frac{a}{b}\)
=> b(a+2016)=a(b+2015)
=>ba+b.2016= ab+a.2015
=>b.2016=a.2015 ( Rút gọn 2 vế với ab)
=>\(\frac{b}{2015}=\frac{a}{2016}\left(đpcm\right)\)
Nếu: \(\frac{a+2016}{a-2016}\)= \(\frac{b+2015}{b-2015}\)
(a + 2016).(b - 2015) = (b + 2015).(a - 2016)
a.b - 2015.a + 2016.b - 2015.2016 = b.a - 2016.b + 2015.a - 2015.2016
2a.2015 = 2b.2016
a.2015 = b.2016
Thì: \(\frac{a}{2016}\)= \(\frac{b}{2015}\)