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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
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)
\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)
\(\Rightarrow x=2013\)
Vậy \(x=2013\)
Tử số = T = \(\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+.....+\frac{1}{2012}\)\(=\left(\frac{2012}{1}+1\right)+\left(\frac{2011}{2}+1\right)+\left(\frac{2010}{3}+1\right)+...+\left(\frac{1}{2012}+1\right)-2012\)\(=2013+\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}-2012\)
\(=\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+1\)
\(=\frac{2013}{2}+\frac{2013}{3}+....+\frac{2013}{2012}+\frac{2013}{2013}\)
\(=2013\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2012}+\frac{1}{2013}\right)\)
Dặt mẫu số là M thì y = \(\frac{2013.M}{M}\)= 2013
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{1+\frac{2012}{2011}+\frac{2012}{2010}+\frac{2012}{2009}+...+\frac{2012}{2}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\frac{2012}{2011}+...+\frac{2012}{2}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{2}\right)}=\frac{1}{2012}\)