So sánh
A= \(\frac{7^{2013}+1}{7^{2015}+1}\)
B= \(\frac{7^{2015}+1}{7^{2017}+1}\)
Những câu hỏi liên quan
Câu 1: so sánh
A = \(\dfrac{7^{2013}+1}{7^{2014}+1}\) và B = \(\dfrac{7^{2014}+1}{7^{2015}+1}\)
Tham khảo: https://hoc24.vn/cau-hoi/so-sanh-afrac720131720141va-bfrac720141720151.235038064198
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cho A=\(\frac{7^{2011}+1}{7^{2013}+1}\)và B=\(\frac{7^{2013}+1}{7^{2015}+1}\)HÃY SO SÁNH A VÀ B
So sanh:
A=72013+1 / 72015+1
B=72015+1 / 72017+1
SO SÁNH \(A=\frac{7^{2013}+1}{7^{2014}+1}\)VÀ \(B=\frac{7^{2014}+1}{7^{2015}+1}\)
\(\frac{A}{B}=\frac{7^{2013}+1}{7^{2014}+1}.\frac{7^{2015}+1}{7^{2014}+1}=\frac{7^{4028}+7^{2013}+7^{2015}+1}{7^{4028}+2.7^{2014}+1}=\)
\(=\frac{7^{4028}+7^{2013}\left(1+7^2\right)+1}{7^{4028}+2.7.7^{2013}+1}=\frac{7^{4028}+50.7^{2013}+1}{7^{4028}+14.7^{2013}+1}>1\)
\(\Rightarrow A>B\)
A/B sao lại nhân v bn
A/B thành A nhân với nghịch đảo của B mà
\(\frac{7^{2015}+1}{7^{2017}+1}\)và\(\frac{7^{2017}+1}{7^{2019}+1}\)
Hãy so sánh
Đặt A= \(\frac{7^{2015}+1}{7^{2017}+1}\)
B= \(\frac{7^{2017}+1}{7^{2019}+1}\)
Ta có A= \(\frac{7^2\left(7^{2015}+1\right)}{7^2\left(7^{2017}+1\right)}\)
= \(\frac{7^{2017}+49}{7^{2019}+49}\)
= \(\frac{7^{2017}+1+48}{7^{2019}+1+48}\)
Vì \(\frac{7^{2017}+1+48}{7^{2019}+1+48}\)>\(\frac{7^{2017}+1}{7^{2019}+1}\)
=> A>B
K MK NHA !
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Bạn tham khảo nhé
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{7^{2017}+1}{7^{2019}+1}< \frac{7^{2017}+1+48}{7^{2019}+1+48}=\frac{7^{2017}+49}{7^{2019}+49}=\frac{7^2\left(7^{2015}+1\right)}{7^2\left(7^{2017}+1\right)}=\frac{7^{2015}+1}{7^{2017}+1}=B\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
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\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
Cho A= \(\frac{4+\frac{4}{2012}-\frac{4}{2013}+\frac{4}{2014}-\frac{4}{2015}}{\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}+7}\)
Và B= \(\frac{1+2+2^2+...+2^{2013}}{2^{2015}-2}\)
Tính A - B
p/S: LM ƠN GIÚP TỚ VS :
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
ĐẶT \(C=1+2+...+2^{2013}\)
\(\Rightarrow2C=2+2^2+...+2^{2014}\)
\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)
\(\Rightarrow C=2^{2014}-2\)
\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)
\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)
\(B=\frac{1}{2}\)
\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)
\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)
VẬY, \(A-B=\frac{-1}{14}\)
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Tính
\(2.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2013}+\frac{1}{2015}+\frac{1}{2017}\right)\)
\(2.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2013}+\frac{1}{2015}+\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(=2.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(=2.\left(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\right)-1\)
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SO SÁNH: A=\(\frac{7^{2013}+1}{7^{2014}+1}\) B=\(\frac{7^{2014}+1}{7^{2015}+1}\)
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