Chứng minh:
\(\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< \frac{5}{6}\)
cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{99.100}\)
CMR: \(̃̃̃̃\frac{7}{12}< A< \frac{5}{6}\)
chứng minh rằng:
\(\frac{7}{12}\)<A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....\frac{1}{99.100}.\)Chứng minh rằng:
a.\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}.\)
b.\(\frac{7}{12}< A< \frac{5}{6}.\)
cho A= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.........+\frac{1}{99.100}\)
CM \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)<\(\frac{3}{4}\)
A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)CM \(\frac{7}{12}< A< \frac{5}{6}\)
cho A =\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
Chứng minh \(\frac{7}{12}< A< \frac{5}{6}\)
------ Các bn giúp mik bài này vs -------
Bài 1 :
a) Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
CMR: \(\frac{7}{12}< A< \frac{5}{6}\)
b) Tìm x , y trái dấu thỏa mãn đề bài : \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y}\)
P= \(\frac{1}{1.2}\) + \(\frac{1}{3.4}\)+ \(\frac{1}{5.6}\)+ ... + \(\frac{1}{99.100}\). Chứng minh : \(\frac{7}{12}\)< P < \(\frac{5}{6}\)