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Co quy luat nay ne em: 1+2=3=2.3:2; 1+2+3=6=3.4:2;...;1+2+3+...+2012=2012.2013:2
Suy ra ta co:
Mau so cua D=1 + 1/(2.3:2) + 1/(3.4:2) + 1/(4.5:2) + .... + 1/(2012.2013:2)
=1 + 2/2.3 + 2/3.4 + 2/4.5 + .... + 2/2012.2013
= 2.[1/2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/2012.2013]
=2.[1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ..... + 1/2012.2013]
=2.[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +....+1/2012 - 1/2013
=2[1 - 1/2013]
=2.2012/2013
Vay D= 2.2012 / (2.2012:2013)=2013
Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)
\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2012\right).2012:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(=2.\left(1-\frac{1}{2013}\right)=\frac{2.2012}{2013}\)
Phân số đề bài cho = \(\frac{2.2012}{\frac{2.2012}{2013}}=2013\)
\(a.\dfrac{-3}{8}-\dfrac{13}{65}+\dfrac{3}{8}=\left(\dfrac{3}{8}-\dfrac{3}{8}\right)-\dfrac{13}{65}=-\dfrac{13}{65}\)
\(b.\left(\dfrac{-13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)=\dfrac{-13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\\ =\left(\dfrac{-13}{7}+\dfrac{10}{7}\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)=-\dfrac{3}{7}\)
\(c.17\dfrac{1}{3}\cdot\left(\dfrac{-3}{7}\right)+3\dfrac{2}{3}\cdot\left(\dfrac{-3}{7}\right)=\dfrac{-3}{7}\cdot\left(17\dfrac{1}{3}+3\dfrac{2}{3}\right)\\ =\dfrac{-3}{7}\cdot\left(\dfrac{52}{3}+\dfrac{11}{3}\right)=\dfrac{-3}{7}\cdot21=-9\)
a: \(=\dfrac{28-2-3}{4}:\dfrac{40-2-5}{8}=\dfrac{23}{4}\cdot\dfrac{8}{33}=\dfrac{46}{33}\)
b: =78(0,65+0,35)+2020(2,2-2,2)
=78*1=78
Sửa đề :
\(A=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2004}}\)
\(2A-A=\left(1+2+\frac{1}{2}+...+\frac{1}{2^{2004}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2005}}\right)\)
\(A=2-\frac{1}{2^{2005}}\)
\(\begin{array}{l}a)A = 32,125 - (6,325 + 12,125) - (37 + 13,675)\\ = 32,125 - 6,325 - 12,125 - 37 - 13,675\\ = (32,125 - 12,125) + ( - 6,325 - 13,675) - 37\\ = 20 + ( - 20) - 37\\ = - 37\\b)B = 4,75 + {\left( {\frac{{ - 1}}{2}} \right)^3} + 0,{5^2} - 3.\frac{{ - 3}}{8}\\ = 4,75 + \frac{{ - 1}}{8} + 0,25 + \frac{9}{8}\\ = (4,75 + 0,25) + \left( {\frac{{ - 1}}{8} + \frac{9}{8}} \right)\\ = 5 + \frac{8}{8}\\ = 5 + 1\\ = 6\\c)C = 2021,2345.2020,1234 + 2021,2345.( - 2020,1234)\\ = 2021,2345.[2020,1234 + ( - 2020,1234)]\\ = 2021,2345.0\\ = 0\end{array}\)
\(\begin{array}{l}\frac{7}{6}.3\frac{1}{4} + \frac{7}{6}.( - 0,25)\\ = \frac{7}{6}.\frac{{13}}{4} + \frac{7}{6}.\frac{{ - 25}}{{100}}\\ = \frac{7}{6}.\frac{{13}}{4} + \frac{7}{6}.\frac{{ - 1}}{4}\\ = \frac{7}{6}.[\left( {\frac{{13}}{4} + ( - \frac{1}{4})} \right)]\\ = \frac{7}{6}.\frac{{12}}{4}\\ = \frac{7}{6}.3\\ = \frac{7}{2}\end{array}\)
\(2+3.\left(\frac{-1}{3}\right)^2\)
=\(2+3.\frac{1}{9}\)
= \(2+\frac{1}{3}\)
=\(\frac{7}{3}\)
= (-1/3.3).(-1/3) + 2 = 1/3 + 2 = 7/3