Tính nhanh:
a) 2+5+8+11+...+104+107.
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
tính nhanh:
B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
A=\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)
B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
B= \(\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
B=\(\frac{3}{4}\)
Sau mình làm tiếp vội quá! k mình nha
Tính nhanh:
a) \(2.\frac{3}{7}+\left(\frac{2}{9}-1\frac{3}{7}\right)-\frac{5}{3}:\frac{1}{9}\)
b) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}\)
c )\(\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)
d) \(19\frac{5}{8}:\frac{7}{12}-15\frac{1}{4}:\frac{7}{12}\)
e) \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{96}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{48}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{24}\)
...
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{3}\Rightarrow A=\frac{2}{3}-\frac{1}{96}=\frac{2\cdot32-1}{96}=\frac{63}{96}=\frac{21}{32}\).
Câu 1: Tính nhanh hợp lí \(\frac{119x83-183}{120x83x-266}\)
Câu 2: Tính nhanh hợp lí: \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
Câu 1
Ta có \(\frac{119x83-183}{120x83-266}=\frac{119x83-183}{119x83+83-266}=\frac{119x83-183}{119x83-183}=1\)
a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\) b) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
c) \(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...............+\frac{1}{1280}\) d) \(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
e) \(1+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}+\frac{5}{64}\) f) \(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
g) \(3+\frac{3}{5}+\frac{3}{25}+\frac{3}{125}+\frac{3}{625}\)
Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?
A) \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
2A= \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
2A-A = \(1-\dfrac{1}{32}\)
A= \(\dfrac{31}{32}\)
b)\(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
Dặt A=
3A= \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
3A-A=\(1-\dfrac{1}{792}\)
2A= \(\dfrac{791}{792}\)
A= \(\dfrac{791}{792}:2=\dfrac{791}{1584}\)
a) Quy đồng mẫu các phân số sau:
i.\(\frac{5}{{12}}\) và \(\frac{7}{{30}}\); ii.\(\frac{1}{2};\,\,\frac{3}{5}\) và \(\frac{5}{8}\).
b) Thực hiện các phép tính sau:
i.\(\frac{1}{6} + \frac{5}{8}\); ii.\(\frac{{11}}{24} - \frac{7}{{30}}\)
a)
i.Ta có: BCNN(12, 30) = 60
60 : 12 = 5; 60 : 30 = 2. Do đó:
\(\frac{5}{{12}} = \frac{{5.5}}{{12.5}} = \frac{{25}}{{60}}\) và \(\frac{7}{{30}} = \frac{{7.2}}{{30.2}} = \frac{{14}}{{60}}.\)
ii.Ta có: BCNN(2, 5, 8) = 40
40 : 2 = 20; 40 : 5 = 8; 40 : 8 = 5. Do đó:
\(\frac{1}{2} = \frac{{1.20}}{{2.20}} = \frac{{20}}{{40}}\)
\(\frac{3}{5} = \frac{{3.8}}{{5.8}} = \frac{{24}}{{40}}\)
\(\frac{5}{8} = \frac{{5.5}}{{8.5}} = \frac{{25}}{{40}}\).
b)
i.Ta có: BCNN(6, 8) = 24
24 : 6 = 4; 24: 8 = 3. Do đó
\(\begin{array}{l}\frac{1}{6} + \frac{5}{8} = \frac{{1.4}}{{6.4}} + \frac{{5.3}}{{8.3}}\\ = \frac{4}{{24}} + \frac{{15}}{{24}} = \frac{{19}}{{24}}.\end{array}\)
ii. Ta có: BCNN(24, 30) = 120
120: 24 = 5; 120: 30 = 4. Do đó:
\(\begin{array}{l}\frac{{11}}{{24}} - \frac{7}{{30}} = \frac{{11.5}}{{24.5}} - \frac{{7.4}}{{30.4}}\\ = \frac{{55}}{{120}} - \frac{{28}}{{120}} = \frac{{27}}{{120}} = \frac{9}{{40}}\end{array}\)
\(\sqrt[2]{4\cdot9\frac{8}{8}+\frac{48\cdot11+5}{1\cdot\frac{814}{5+\frac{6145}{1\cdot\frac{821}{614}}}}}2548-\frac{8452}{14\cdot\frac{58}{96\cdot\frac{41}{\frac{24}{1\cdot\frac{975545}{1421+\frac{84874}{\frac{1+2+3+4+5+6+7+8+9\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{2\cdot\frac{2}{1}}}}}}}}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+\(\frac{1}{24}\)+\(\frac{1}{48}\)+\(\frac{1}{96}\)
= ( \(\frac{1}{3}\)+\(\frac{1}{6}\)) + ( \(\frac{1}{12}\)+ \(\frac{1}{24}\)) + ( \(\frac{1}{48}\)+\(\frac{1}{96}\))
= \(\frac{1}{2}\) + \(\frac{1}{8}\) + \(\frac{1}{32}\)
= \(\frac{5}{8}\) + \(\frac{1}{32}\)
= \(\frac{21}{32}\)
tính nhanh hộ mình câu này với : (có kèm bài làm nha )
\(\frac{3}{2}+\frac{ 3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(5+\frac{5}{3}+\frac{5}{9}+\frac{5}{27}+\frac{5}{81}\)
\(\frac{4}{3}+\frac{1}{3}+\frac{1}{12}+\frac{1}{48}+\frac{1}{192}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
a)=768/512+192/512+48/512+12/512+3/512
=768+192+48+12+3/512
=1023/512
b)=405/81+135/81+45/81+15/81+5/81
=405+135+45+15+5/81
=595/81
c)=256/192+64/192+16/192+4/192+1/192
=256+64+16+4+1/192
=341/192