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21 tháng 3 2015

T/c:A=1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+...+1/97*98*99+1/98*99*100

2A=2/1*2*3+2/2*3*4+2/3*4*5+2/4*5*6+...+2/97*98*99+1/98*99*100

2A=(1/1*2-1/2*3)+(1/2*3-1/3*4)+(1/3*4-1/4*5)+.....+(1/97*98-1/98*99)+(1/98*99-1/99*100)

2A=1/2+1/99*100

A=tự tính nha

19 tháng 2 2018

A= [(1/2-1/2*3)/2]+[(1/2-1/3*4)/2]+...+[(1/2-1/99*100)/2]

A=(1/2-1/99*100)/2

A=-101/198/2

A=-101/396

16 tháng 9 2015

A=(98.99.100.101-0.1.2.3):4=242550

8 tháng 6 2017

=348450 nha bạn

2 tháng 2 2016

24 , ủng hộ mk nha

2 tháng 2 2016

\(\frac{a}{b}=\frac{1}{3}\) suy ra 3a=b

thay vào biểu thức A ta có :

A=\(\frac{9a+5b}{b-2a}=\frac{3b+5b}{b-\frac{2}{3}b}=\frac{8b}{\frac{1}{3}b}=8b\div\frac{1}{3}b=8b\times\frac{3}{b}=24\)

1 tháng 7 2015

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{38.39}\right)=\frac{185}{741}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}\left(\frac{741}{1482}-\frac{1}{1482}\right)\)

\(=\frac{1}{2}.\frac{370}{741}\)

\(=\frac{185}{741}\).

+)TH1:a=\(\frac{1}{3}\) và b=\(\frac{1}{4}\)

\(y=\frac{5}{3}.\frac{1}{3}-\frac{3}{\frac{1}{4}}=-\frac{67}{6}\)

+)TH2 a=\(-\frac{1}{3}\)và b=\(-\frac{1}{4}\)

\(y=\frac{5}{3}.-\frac{1}{3}-\frac{3}{-\frac{1}{4}}=\frac{103}{9}\)

8 tháng 3 2015

b)\(\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}}\)

\(=\frac{2.2014}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2014.2015}{2}}}\)

\(=\frac{2.2014}{2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)}\)

\(=\frac{2014}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}}\)

\(=\frac{2014}{1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}}\)

\(=\frac{2014}{1-\frac{1}{2015}}=\frac{2014}{\frac{2014}{2015}}=2015\)

8 tháng 3 2015

a)\(\frac{\frac{105}{13.20}+\frac{105}{20.27}+...+\frac{105}{62.69}}{\frac{5}{9.13}+\frac{7}{9.25}+\frac{13}{19.25}+\frac{31}{19.69}}\)

\(=\frac{\frac{105}{7}.\left(\frac{7}{13.20}+\frac{7}{20.27}+...+\frac{7}{62.69}\right)}{\left(\frac{9}{9.13}-\frac{4}{9.13}\right)+\left(\frac{16}{9.25}-\frac{9}{9.25}\right)+\left(\frac{19}{19.25}-\frac{6}{19.25}\right)+\left(\frac{50}{19.69}-\frac{19}{19.69}\right)}\)

\(=\frac{\frac{105}{7}.\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+...+\frac{1}{62}-\frac{1}{69}\right)}{\left(\frac{1}{13}-\frac{1}{9}+\frac{1}{13}\right)+\left(\frac{1}{9}-\frac{1}{25}-\frac{1}{25}\right)+\left(\frac{1}{25}-\frac{1}{19}+\frac{1}{25}\right)+\left(\frac{1}{19}-\frac{1}{69}-\frac{1}{69}\right)}\)

\(=\frac{\frac{105}{7}.\left(\frac{1}{13}-\frac{1}{69}\right)}{\frac{1}{13}+\frac{1}{13}-\frac{1}{69}-\frac{1}{69}}=\frac{\frac{105}{7}\left(\frac{1}{13}-\frac{1}{69}\right)}{2\left(\frac{1}{13}-\frac{1}{69}\right)}=\frac{\frac{105}{7}}{2}=\frac{15}{2}\)