Các câu hỏi tương tự
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 = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Tìm x biết:
a) \(^{2^x+2^{x+1}+2^{x+2}+2^{x+3}=480}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right).x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{2}{2011}+\frac{1}{2012}\)
Bài \(1:\)TÌM \(x:\)
\(a,\) \(x.\)\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}{\frac{2011}{1}+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{2}{2011}+\frac{1}{2012}}=1\)
bài 1 :a) Tính M:\(\frac{\frac{7}{2012}+\frac{7}{9}-\frac{1}{4}}{\frac{5}{9}-\frac{3}{2012}-\frac{1}{2}}\)
b) So sánh A và B biết A =\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\);;; B =\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{17}\)
Bài 3 : a) Tính
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) Tính :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\frac{1}{2011}}\)
Tìm x biết:
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)\cdot x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(y=\frac{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\) =?
Bài 1:TÌM x:a,frac{1}{3}+frac{1}{6}+frac{1}{10}+...+frac{2}{x.left(x+1right)}frac{2}{2013}b,frac{x+4}{2000}+frac{x+3}{2001}frac{x+2}{2002}+frac{x+1}{2003}c,frac{x+5}{205}+frac{x+4}{204}+frac{x+3}{203}frac{x+166}{366}+frac{x+167}{367}+frac{x+168}{368} d, x.frac{frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2011}+frac{1}{2012}}{frac{2011}{1}+frac{2011}{2}+frac{2010}{3}+frac{2009}{4}+...+frac{2}{2011}+frac{1}{2012}}1
Đọc tiếp
Bài \(1:\)TÌM \(x:\)
\(a,\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}\)
\(b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(c,\frac{x+5}{205}+\frac{x+4}{204}+\frac{x+3}{203}=\frac{x+166}{366}+\frac{x+167}{367}+\frac{x+168}{368}\)
\(d,\) \(x.\)\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}{\frac{2011}{1}+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{2}{2011}+\frac{1}{2012}}=1\)
Chứng tỏ rằng A<1 biết
A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{2010^2}+\frac{1}{2011^2}+\frac{1}{2012^2}<1\)