\(B=\frac{1}{1.4}+\frac{1}{4.3}+\frac{1}{3.8}+...+\frac{1}{7.16}+\frac{1}{16.9}+\frac{1}{9.20}\)
tìm B
Những câu hỏi liên quan
Chứng minh : \(\frac{1}{1.4}+\frac{1}{3.8}+\frac{1}{5.12}+....+\frac{1}{99.200}<\frac{5}{12}\)
Tính:
a, \(2015.20142014-2014.20152014\)
b, \(2^2+4^2+6^2+8^2+..+30^2\)
c,\(\frac{1}{4.3}+\frac{1}{6.4}\frac{1}{8.5}\frac{1}{10.6}\frac{1}{12.7}\frac{1}{14.8}\frac{1}{16.9}\frac{1}{18.10}\)
Ai làm nhanh và giải đầy đủ nhất thì mik tik cho
\(\frac{1}{1.4}+\frac{1}{2.6}+\frac{1}{3.8}+...+\frac{1}{20.42}\)
Tính giá trị biểu thức trên.
Lưu ý: Dấu chấm(.) là dấu nhân
Đây là toán lớp 6 nha
Chứng minh rằng \(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)<1.5
\(\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{1}{2012.1342}\)
\(=\frac{3}{1.4}+\frac{3}{2.6}+\frac{3}{3.8}+...+\frac{3}{2012.4026}\)
\(=\frac{6}{2.4}+\frac{6}{4.6}+\frac{6}{4.8}+...+\frac{6}{4024.4026}\)
\(=3\cdot\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{4024.4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=3\cdot\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=3\cdot\frac{1}{2}-3\cdot\frac{1}{4026}\)
\(=1,5-\frac{3}{4026}< 1,5\)
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a, \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
b,\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
c,\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+....+\frac{2}{97.100}\)
a) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}\)
\(=\frac{2.1}{1}=2\)
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b) \(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5}-\frac{3}{10}\)
\(=\frac{0,125-\frac{1}{5}+\frac{1}{7}}{3\left(0,125-\frac{1}{5}+\frac{1}{7}\right)}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)
\(=\frac{1}{3}+\frac{\frac{30}{60}+\frac{20}{60}-\frac{12}{60}}{\frac{45}{60}+\frac{30}{60}-\frac{9}{60}}\)
\(=\frac{1}{3}+\frac{\frac{19}{30}}{\frac{11}{10}}\)
\(=\frac{1}{3}+\frac{19}{33}=\frac{11}{33}+\frac{19}{33}\)
\(=\frac{30}{33}=\frac{10}{11}\)
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Xem thêm câu trả lời
Tìm xa, (frac{11}{1.3} + frac{11}{3.5}+ ...... + frac{11}{97.99}) . 2x frac{49}{9}b, frac{1}{1.4} + frac{1}{4.7}+.......+ frac{1}{97.100} frac{0,33.x}{2009}c, (frac{1}{3.8}+ frac{1}{8.13}+ ........ + frac{1}{103.108}) . 540 - 2x 14
Đọc tiếp
Tìm x
a, (\(\frac{11}{1.3}\) + \(\frac{11}{3.5}\)+ ...... + \(\frac{11}{97.99}\)) . 2x = \(\frac{49}{9}\)
b, \(\frac{1}{1.4}\) + \(\frac{1}{4.7}\)+.......+ \(\frac{1}{97.100}\)= \(\frac{0,33.x}{2009}\)
c, (\(\frac{1}{3.8}\)+ \(\frac{1}{8.13}\)+ ........ + \(\frac{1}{103.108}\)) . 540 - 2x = 14
Cho biểu thứcA= \(\frac{\frac{1}{3.8}+\frac{1}{8.13}+...+\frac{1}{33.38}}{\frac{21}{3.10}+ \frac{15}{10.15}+\frac{27}{15.24}+\frac{9}{24.27}+\frac{33}{27.38}}\)
a,Tính A
b,So snah A và B với B=\(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{81}+\frac{1}{100}\)
Chứng minh rằng:A=\(\frac{3}{1.4}\)+\(\frac{3}{2.6}\)+\(\frac{3}{3.8}\)+....+\(\frac{1}{2012.1342}\)<1;5
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...............+\dfrac{1}{2012.1342}\)
\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...........................+\dfrac{3}{2012.4026}\)
\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+..........................+\dfrac{6}{4024.4026}\)
\(A=3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...................+\dfrac{2}{4024.4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+....................+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
\(A=3.\dfrac{1}{2}-3.\dfrac{1}{4026}\)
\(A=1,5-\dfrac{3}{4026}< 1,5\)
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Ta có
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{1}{2012.1342}\)
A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{3}{2012.4026}\)
A = \(\dfrac{6}{2.4}\) + \(\dfrac{6}{4.6}\) + \(\dfrac{6}{6.8}\) + ... + \(\dfrac{6}{4024.4026}\)
A = \(3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{4024.4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)
A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)
A = 3.\(\dfrac{1}{2}\) - 3.\(\dfrac{1}{4026}\)
A = 1,5 - \(3.\dfrac{1}{4026}\) < 1,5
=> A < 1,5
=> đpcm
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Cô @Bùi Thị Vân hình như có gì đó nhầm lẫn!!
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Chứng minh rằng:A=\(\frac{3}{1.4}\)+\(\frac{3}{2.6}\)+\(\frac{3}{3.8}\)+....+\(\frac{1}{2012.1342}\)<1;5
\(A=\)\(\frac{3}{1.4}\)\(+\)\(\frac{3}{2.6}\)\(+\)\(\frac{3}{2.8}\)\(+\).........\(+\)\(\frac{1}{2012.1342}\)\(< 1,5\)
\(=\)\(\frac{3}{1.4}\)\(+\)\(\frac{3}{2.6}\)\(+\)\(\frac{3}{3.8}\)\(+\)............\(+\)\(\frac{3}{2012.4026}\)
\(=\)\(\frac{6}{2.4}\)\(+\)\(\frac{6}{4.6}\)\(+\)\(\frac{6}{6.8}\)\(+\)..............\(+\)\(\frac{6}{4024.4026}\)
\(=\)\(3.\)\(\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...........+\frac{2}{4024.4026}\right)\)
\(=\)\(3.\)\(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{4024}-\frac{1}{4026}\right)\)
\(=\)\(3.\)\(\left(\frac{1}{2}-\frac{1}{4026}\right)\)
\(=\)\(3.\)\(\frac{1}{2}\)\(-\)\(3.\)\(\frac{1}{4026}\)
\(=\)\(1,5\)\(-\)\(\frac{3}{4026}\)\(< \)\(1,5\)
Vậy \(A< 1,5\)
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