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\(3C=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{27.28.29.30}\)
\(3C=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{30-27}{27.28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{27.28.29}+\frac{1}{28.29.30}\)
\(3C=\frac{1}{1.2.3}-\frac{1}{28.29.30}\Rightarrow C=\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right):3\)
\(\frac{-3}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}\)
\(=\frac{-3}{5}.\frac{4}{7}+\frac{-3}{5}.\frac{2}{7}+\frac{-3}{5}.\frac{1}{7}\)
\(=\frac{-3}{5}\left(\frac{4}{7}+\frac{2}{7}+\frac{1}{7}\right)\)
\(=\frac{-3}{5}.1\)
\(=\frac{-3}{5}\)
\(-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}x\frac{2}{7}+-\frac{3}{5}=-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}\left(\frac{2}{7}+1\right)=-\frac{2}{5}x\frac{1}{7}+-\frac{3}{5}x\frac{9}{7}.\)
\(-\frac{2}{35}-\frac{27}{35}=-\frac{29}{35}\)
tôi tính nhẩm vậy ko biết có đứng ko
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)
\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)
B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)
= \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)
= \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)
= \(\frac{-22}{105}\)
C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}=\frac{6}{7}\)
\(E=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(E=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(E=2\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)
\(E=2\left(\frac{1}{4x5}+\frac{1}{5x6}+...+\frac{1}{15x16}\right)\)
\(E=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(E=2\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(E=\frac{3}{8}\)
1/2E=1/20+1/30+1/42+...+1/240. =>1/2E=1/4*5+1/5*6+1/6*7+...+1/15*16. =>1/2E=1/4-1/5+1/5-1/6+1/6-1/7+...+1/15-1/16. =>1/2E=1/4-1/16=3/16. =>E=3/16:1/2=3/8. Câu b có vấn đề.
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))
=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))
=>A=\(\frac{350}{101}\)
7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))
7/2 ( 1 - 1/101 )
7/2 x 100/101
=350/101
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
= 1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - 1 / 6
Ta gạch các ps trùng.
Còn lại :
1 / 1 - 1 / 6 = 6 / 5
\(\frac{2}{7}.5\frac{1}{4}-\frac{2}{7}.3\frac{1}{4}\)
= \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
= \(\frac{2}{7}.2\)
= \(\frac{4}{7}\)
\(=\frac{2}{7}\times\left(5\frac{1}{4}-3\frac{1}{4}\right)=\frac{2}{7}\times2=\frac{4}{7}\)