A= \(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+....+\(\dfrac{1}{39800}\)
Tính (theo mẫu).
Mẫu: \(\dfrac{1}{2}-\dfrac{5}{12}=\dfrac{6}{12}-\dfrac{5}{12}=\dfrac{6-5}{12}=\dfrac{1}{12}\) |
a) \(\dfrac{3}{4}-\dfrac{1}{8}\) b) \(\dfrac{2}{6}-\dfrac{5}{18}\) c) \(\dfrac{2}{5}-\dfrac{3}{20}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
Tính:
\(a,\left(\dfrac{5}{12}-\dfrac{3}{4}\right).\dfrac{-6}{5}+\dfrac{-7}{6}\)
\(b,\left(5\dfrac{1}{2}-5\right)^2+\dfrac{-1}{6}:\dfrac{5}{3}\)
\(c,75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
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) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)
tìm x bt
a) \(\dfrac{2}{3}\)x - \(\dfrac{7}{6}\) = \(\dfrac{12}{7}\) - \(\dfrac{1}{2}\)
b) ( \(1\dfrac{1}{2}\) + \(\dfrac{5}{3}\) - \(\dfrac{1}{6}\) ) : x = \(\dfrac{3}{4}\) - \(\dfrac{1}{2}\)
Lời giải:
a.
$\frac{2}{3}x-\frac{7}{6}=\frac{12}{7}-\frac{1}{2}=\frac{17}{14}$
$\frac{2}{3}x=\frac{17}{14}+\frac{7}{6}=\frac{50}{21}$
$x=\frac{50}{21}: \frac{2}{3}=\frac{25}{7}$
b.
$(1\frac{1}{2}+\frac{5}{3}-\frac{1}{6}):x=\frac{3}{4}-\frac{1}{2}$
$3:x=\frac{1}{4}$
$x=3: \frac{1}{4}=12$
a) \(\dfrac{1}{2}+\dfrac{-1}{-3}-\dfrac{5}{12}< 2x< \dfrac{12}{-31}+\dfrac{136}{31}\)
b) \(\dfrac{-2}{5}< \dfrac{x}{15}< \dfrac{1}{6}\)
HéP :)
`a)1/2+[-1]/[-3]-5/12 < 2x < 12/[-31]+136/31`
`186/372+124/372-155/372 < [744x]/372 < [-144]/372+1632/372`
`186+124-155 < 744x < -144+1632`
`155 < 744x < 1488`
`155:744 < 744x:744 < 1488:744`
`5/24 < x < 2`
Vậy `5/24 < x < 2`
__________________________________________________
`b)[-2]/5 < x/15 < 1/6`
`[-12]/30 < [2x]/30 < 5/30`
`-12 < 2x < 5`
`-12:2 < 2x:2 < 5:2`
`-6 < x < 5/2`
Vậy `-6 < x < 5/2`
Giải:
a) x - \(\dfrac{9}{25}\)= \(\dfrac{16}{25}\)
x = \(\dfrac{16}{25}\)+\(\dfrac{9}{25}\)
x = \(\dfrac{25}{25}\)
x = 1
b) \(\dfrac{-12}{30}\)<\(\dfrac{x}{30}\)<\(\dfrac{5}{30}\)
=> x có thể bằng \(\dfrac{-11}{30}\) đến \(\dfrac{4}{30}\)
=> x bằng -5; -4; -3; -2; -1;0;1;2
a=(\(\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)):(\(\dfrac{-6}{51}-\dfrac{6}{52}-\dfrac{6}{53}-...-\dfrac{6}{100}\))
giúp mik giải nhé
cảm ơn !
Tính: \(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(A=1-\dfrac{1}{8}=\dfrac{7}{8}\)
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(A=\dfrac{367}{420}\approx0,87\)
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1/1 − 1/2 + 1/2 − 1/3 + 1/3 − 1/4 + 1/4 − 1/5 + 1/5 − 1/6 + 1/6 − 1/7 + 1/7 − 1/8 + 1/8 − 1/9
= 1 − 1/9
= 8/9
Tính
a,\(\dfrac{3}{4}\) - \(\dfrac{1}{12}\) - \(\dfrac{1}{6}\) + 2 b,\(\dfrac{2}{5}\) + \(\dfrac{3}{10}\) + \(\dfrac{4}{5}\) - \(\dfrac{5}{6}\)
c, \(\dfrac{12}{25}\) : \(\dfrac{9}{32}\) \(\dfrac{16}{15}\) d, \(\dfrac{3}{5}\) x \(\dfrac{2}{7}\) x \(\dfrac{5}{6}\) : 5
a: =9/12-1/12-2/12+2=1/2+2=5/2
b: =(2/5+4/5+3/10)-5/6=6/5+3/10-5/6=15/10-5/6=3/2-5/6=9/6-5/6=4/6=2/3
c: \(=\dfrac{12}{25}\cdot\dfrac{32}{9}\cdot\dfrac{15}{16}=\dfrac{12}{9}\cdot\dfrac{32}{16}\cdot\dfrac{15}{25}=\dfrac{4}{3}\cdot\dfrac{3}{5}\cdot2=\dfrac{8}{5}\)
VÒNG 2
Bài 1: Mèo con nhanh nhẹn
\(\dfrac{1}{2}\) + \(\dfrac{1}{12}\) | 2 + \(\dfrac{1}{6}\) | \(\dfrac{1}{20}\) | 1 - \(\dfrac{1}{9}\) | |
\(\dfrac{1}{15}\) + \(\dfrac{2}{15}\) | \(\dfrac{1}{2}\) + \(\dfrac{2}{3}\) | \(\dfrac{7}{12}\) | \(\dfrac{4}{12}\) | |
\(\dfrac{9}{14}\)+ \(\dfrac{1}{14}\) | 1 + \(\dfrac{1}{6}\) | \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) | \(\dfrac{1}{3}\) - \(\dfrac{2}{9}\) | |
\(\dfrac{3}{2}\) + \(\dfrac{2}{3}\) | \(\dfrac{1}{5}\) | 1 - \(\dfrac{8}{9}\) | ||
\(\dfrac{5}{7}\) | 1 - \(\dfrac{2}{3}\) | \(\dfrac{1}{3}\) + \(\dfrac{5}{9}\) |