Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2015}{2016}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2013}{2014}\)
\(\Rightarrow A>\frac{1.2.3...2013}{2.3.4...2014}\)
\(\Rightarrow A>\frac{1}{2014}>\frac{1}{2017}\)
Vậy \(A>\frac{1}{2017}\left(đpcm\right)\)
ta có 2015/2016+2016/2017+2017/2015=(1-1/2016)+(1-1/2017)+(2+1/2015)
=4-(1/2016+1/2017-1/2015)
1/2016<1; 1/2017<1 nên 1/2016+1/2017<2 suy ra 1/2016+1/2017-1/2015<1(vì 1/2015<1)
4-(1/2016+1/2017-1/2015)>4-1=3
2015/2016+2016/2017+2017/2015>3
cho mik nhé
\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
câu 1. tìm x nguyên để \(\frac{-35}{6}\)<x<\(\frac{-18}{5}\)
<=> -4,375<x<-3,6
mà x\(\in\)Z nên x={-4}
câu 2. A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)
Vì \(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)
Vậy B<A
\(A=\frac{2015}{2016}+\frac{2016}{2017}=1-\frac{1}{2016}+1-\frac{1}{2017}>1\)
\(B=\frac{2015+2016}{2016+2017}< \frac{2016+2017}{2016+2017}=1\)
Suy ra \(A>B\).
Mấy bài dạng này biết cách làm là oke
Ta có :
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\left(2016-1-1-...-1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(A=2017\)
Vậy \(A=2017\)
Chúc bạn học tốt ~
\(A=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2016+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
(số 2016 tách ra làm 2016 số 1 rồi cộng vào từng phân số, còn dư 1 số viết thành 2017/2017 nghe bạn!!! :)))
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}\)
\(A=2017\)
