Rút gọn:
\(F=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+........+\frac{2}{2010}+\frac{1}{2011}}\)
Những câu hỏi liên quan
\(S=\sqrt{1+2010^2+\frac{2010^2}{2011^2}}+\frac{2010}{2011}+\sqrt{1+2011^2+\frac{2011^2}{2012^2}}+\frac{2011}{2012}+\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}\)
So sánh P và Q biết : P = 2010/2011 + 2011/2012 + 2012/2013 và Q = 2010+2011+2012/ 2011 +2012+2013
Chứng tỏ N < 1 với N = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}
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\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
P=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+.....+\frac{1}{2011}}\)
So sánh : \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2011}+\frac{2012}{2010}}\) và \(\frac{2016}{2017}\)
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 \(\frac{1}{1+\frac{2010}{2011}+\frac{2010}{2012}}+\frac{1}{1+\frac{2011}{2010}+\frac{2011}{2012}}+\frac{1}{1+\frac{2012}{2010}+\frac{2012}{2011}}>\frac{2016}{2017}\)
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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
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mình ko hiểu lắm sao tự nhiên lại đang \(\frac{1}{2010.\left[2010+2011+2012\right]}\)lại sang luôn \(\frac{\frac{1}{2010}}{2010+2011+2012}\)
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Xem thêm câu trả lời
Tính C=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2011+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)}\)
\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}}\)
\(C=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
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TÍNH : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}}\)
nâng cao phát triển toán 7 đấy
mấy bài đấu thì phải
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Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)
\(L=1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(L=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}\)
\(L=2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2012}\right)\)
Hay: \(P=\frac{1}{2012}\)
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Tính P:
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
Ta có :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2012}\right)}\)
\(\frac{1}{2012}\)
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\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{2011}+\frac{1}{2012}\)
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\cdot\cdot\cdot\frac{2}{2010}+\frac{1}{2011}\)
Tính \(\frac{B}{A}\)
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
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từ dề bài ta có :
\(\frac{B}{A}=\frac{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}\)
\(\frac{B}{A}=2012\)
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