Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2011}+\frac{2012}{2010}}\)
\(=\frac{1}{2010\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)}+\frac{1}{2011\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2012}\right)}+\frac{1}{2012\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)}\)
\(=\frac{\frac{1}{2010}}{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}+\frac{\frac{1}{2011}}{\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2012}}+\frac{\frac{1}{2012}}{\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}{\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}}=1\)
Mà \(\frac{2016}{2017}< 1\)
Vậy \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2010}+\frac{2012}{2011}}>\frac{2016}{2017}\)
dấu cần điền là : >
Vì kết quả của phép tính vế thứ 1 là 1
và phân số 2016/2017 bé hơn 1 nên ta điền dấu lớn
Ta có:
2010.A=\(\frac{2010^{2012}+2010}{2010^{2012}+1}\)
2010.B=\(\frac{2010^{2011}+2010}{2010^{2011}+1}\)
2010.A có phần thừa với 1 là:\(\frac{2009}{2010^{2012}+1}\)
2010.B có phần thừa với 1 là:\(\frac{2009}{2010^{2011}+1}\)
Vì \(\frac{2009}{2010^{2012}+1}
\(A=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\)
\(A=\frac{4064340600}{4066362660}+\frac{4064341605}{4066362660}+\frac{4070408792}{4066362660}\)
\(A=3,000000742\)
\(B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{17}\)
\(B=1,939552553\)
vì đây là so sánh hai dòng phân số nên ta đổi ra thập phân nhé
do 3,000000742 > 1,939552553 và 3 > 1 Nên A > B nhé
đúng thì k nhé
chúc học giỏi !!!!
so sánh : cho A\(\frac{2010^{2011}+1}{2010^{2012}+1}\)
cho B =\(\frac{2010^{2010}+1}{2010^{2011}+1}\)
Ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}\)
\(2010A=\frac{2010^{2012}+2010}{2010^{2012}+1}\)
\(2010A=1+\frac{2009}{2010^{2012}+1}\)
Lại có:
\(B=\frac{2010^{2010}+1}{2010^{2011}+1}\)
\(2010B=\frac{2010^{2011}+2010}{2010^{2011}+1}\)
\(2010B=1+\frac{2009}{2010^{2011}+1}\)
Vì \(1+\frac{2009}{2010^{2012}+1}< 1+\frac{2009}{2010^{2011}+1}\)
nên 2010A < 2010B
hay A < B
Vậy A < B
\(1-A=1-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010^{2012}+1}{2010^{2012}+1}-\frac{2010^{2011}+1}{2010^{2012}+1}=\frac{2010}{2010^{2012}+1}\)
\(1-B=1-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010^{2011}+1}{2010^{2011}+1}-\frac{2010^{2010}+1}{2010^{2011}+1}=\frac{2010}{2010^{2011}+1}\)
Do \(\frac{2010}{2010^{2012}+1}B\)
Do 20102011+1<20102012+1=>A<1
Tương tự với B;B<1
Theo đề bài ta có:
\(A=\frac{2010^{2011}+1}{2010^{2012}+1}
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)
\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)
\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
