Tính :
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
B=\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\)
(Các bn giúp mik vs mai mik phải nộp bài r )
Tính
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\)
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\\)
\(3B=3\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\)
\(3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\)
\(2B=3B-B\)
\(2B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{729-1}{972}=\dfrac{728}{972}=\dfrac{182}{243}\)
\(B=\dfrac{182}{243}:\dfrac{1}{2}=\dfrac{182\cdot2}{243}=\dfrac{364}{243}\)
Bài 2: Tính
D =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{29}\)
E = \(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{972}\)
Tính (Tính hợp lí nếu có thể)
a) \(\dfrac{-7}{12}\)-\(\dfrac{3}{36}\)
b) (4-\(\dfrac{5}{12}\)):2+\(\dfrac{5}{24}\)
c) \(\dfrac{8}{9}\)+\(\dfrac{1}{9}\).\(\dfrac{2}{13}\)+\(\dfrac{1}{9}\).\(\dfrac{11}{13}\)
d) \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}\). ... .\(\dfrac{9999}{10000}\)
e) \(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
*Lưu ý: Mong các anh chị trình bày chi tiết để em có thể hiểu bài, em xin các anh chị đừng viết mỗi kết quả xong em chả biết một cái gì ;-;
a: =-21/36-3/36=-24/36=-2/3
b: =43/12*1/2+5/24=43/24+5/24=2
c: =8/9+1/9=1
e: =1-1/4+1/4-1/7+...+1/97-1/100
=1-1/100=99/100
Tính tổng sau bằng cách hợp lí
A = \(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+...+\dfrac{1}{972}+\dfrac{1}{2916}\)
A=14+112+136+...+1972+12916
3A=34+14+112+...+1324+1972
3A−A=(34+14+112+...+1324+1972)−(14+112+136+...+1972+12916)
2A=34−12916
A=10932916
Tính hợp lí :
A = \(\dfrac{-2}{9}\) + \(\dfrac{-3}{4}\) + \(\dfrac{3}{5}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{57}\) + \(\dfrac{1}{3}\) + \(\dfrac{-1}{36}\)
B = \(\dfrac{1}{2}\) + \(\dfrac{-1}{5}\) + \(\dfrac{-5}{7}\) + \(\dfrac{1}{6}\) + \(\dfrac{-3}{35}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{41}\)
C = \(\dfrac{-1}{2}\) + \(\dfrac{3}{5}\) + \(\dfrac{-1}{9}\) + \(\dfrac{1}{127}\) + \(\dfrac{-7}{18}\) + \(\dfrac{4}{35}\) + \(\dfrac{2}{7}\)
Tính (theo mẫu).
Mẫu: \(2+\dfrac{1}{6}=\dfrac{12}{6}+\dfrac{1}{6}=\dfrac{13}{6};1-\dfrac{1}{4}=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\) |
a) \(1+\dfrac{4}{9}\) b) \(5+\dfrac{1}{2}\) c) \(3-\dfrac{5}{6}\) d) \(\dfrac{31}{7}-2\)
a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
Tính hợp lý:
\(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\) \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
Bài 1 : Thực hiện phép tính ( tính hợp lý nếu có thể )
a ) \(\dfrac{1}{12}+\dfrac{3}{4}\)
b ) \(\dfrac{-4}{7}.1\dfrac{1}{2}\)
c )\(\dfrac{7}{9}+\left(\dfrac{2}{3}+\dfrac{-7}{9}\right)\)
d )\(\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}\)
e )\(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}\)
g )\(2\dfrac{2}{5}.0,25-\left(\dfrac{11}{20}+75\%\right):\dfrac{13}{5}\)
a. 1/12 + 3/4
= 1/12+ 9/12
= 10/12 = 5/6
b. -4/7. 1 1/2
= -4/7. 3/2
= -6/7
c. 7/9+ ( 2/3 + -7/9)
= 7/9 + -1/9
= 8/9 = 2/3
d. 2/3 - 1/3 : 3/4
= 2/3 - 4/9
= 6/9 - 4/9
= 2/9
e. -7/25 . 11/13+ - 7/25. 2/13
=-77/ 325+ -14/ 325
= - 7/25
g. 2 2/5 . 0,25- ( 11/20+ 75%) : 13/5
= 12/5 . 1/4 - ( 11/20 + 3/4) : 13/5
= 12/5 . 1/4 - 13/10 : 12/ 5
= 3/5 - 13/24
= 7/120
lm đại vậy thôi nha hơi lâu vì mik bị lag nên phải làm lại
a) A= \(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
b) B=\(\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
c) C=\(\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-......-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)