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=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{8}+...+\frac{2}{28}-\frac{2}{31}\right)\)
=\(\frac{1}{3}\times\left(\frac{2}{1}-\frac{2}{31}\right)=\frac{20}{31}\)
Bấm đúng cho tui, đi mà. CHÚC BẠN HỌC GIỎI
bài giải đó là sai giả như vầy nè
\(=\frac{1}{3}\cdot2\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{28}-\frac{1}{31}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{31}\right)\)
=\(\frac{2}{3}\cdot\frac{30}{31}=\frac{20}{31}\)
\(B=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+........+\frac{1}{27.30}\right)\)
\(B=3.\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.......-\frac{1}{27}+\frac{1}{27}-\frac{1}{30}\right)\)
\(B=1.\left(\frac{1}{1}-\frac{1}{30}\right)\)
\(B=\frac{29}{30}\)
B =\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
B = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{27}-\frac{1}{30}\)
B =\(\frac{1}{1}-\frac{1}{30}\)
B =\(\frac{29}{30}\)
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{197.200}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{197}-\frac{1}{200}\)
\(=1-\frac{1}{200}\)
\(=\frac{199}{200}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{197}-\frac{1}{200}\)
\(A=1-\frac{1}{200}\)
\(A=\frac{199}{200}\)
=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}\left(\frac{1}{7}-\frac{1}{10}\right)+..........+\frac{1}{3}\left(\frac{1}{97}-\frac{1}{100}\right)\)
=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+........+\frac{1}{97}-\frac{1}{100}\right)\)
=\(\frac{1}{3}\left(\frac{1}{4}-\frac{1}{100}\right)\)
=\(\frac{1}{3}x\frac{6}{25}\)=\(\frac{2}{25}\)
vậy biểu thức trên có giá trị bằng\(\frac{2}{25}\)
a,1/1-1/4+1/4-1/7+...+1/2008-1/2011
=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)
=1-1/2011+0+...+0
=1-1/2011
=2010/2011
a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)
=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))
= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))
=9(1-\(\frac{1}{100}\))
A=\(\frac{891}{100}\)
b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))
=1-\(\frac{1}{30}\)
B=\(\frac{29}{30}\)
a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Đến đây ta suy được ra S<1
Ta có :
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy \(S< 1\)