Tính:a) 1+1/4+1/8+1/16
b)2-1/8-1/12-1/16
c)4/99*18/5:12/11+3/5
d)(1-3/4)*(1+1/3):(1-1/3)
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Những câu hỏi liên quan
KV
21 tháng 10 2023
a) 1 + 1/4 + 1/8 + 1/16
= 16/16 + 4/16 + 2/16 + 1/16
= 23/16
b) 2 - 1/8 - 1/12 - 1/16
= 96/48 - 6/46 - 4/48 - 3/48
= 83/48
c) 4/99 × 18/5 : 12/11 + 3/5
= 8/55 : 12/11 + 3/5
= 2/15 + 3/5
= 2/15 + 9/15
= 11/15
d) (1 - 3/4) × (1 + 1/3) : (1 - 1/3)
= 1/4 × 4/3 : 2/3
= 1/3 : 2/3
= 2
PD
1
H9
HT.Phong (9A5)
CTVHS
18 tháng 10 2023
\(1+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{5}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
\(=\dfrac{11}{8}+\dfrac{1}{16}\)
\(=\dfrac{23}{16}\)
______
\(2-\dfrac{1}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{15}{8}-\dfrac{1}{12}-\dfrac{1}{16}\)
\(=\dfrac{43}{24}-\dfrac{1}{16}\)
\(=\dfrac{83}{48}\)
_________
\(\dfrac{4}{99}\times\dfrac{18}{5}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}:\dfrac{12}{11}+\dfrac{3}{5}\)
\(=\dfrac{8}{55}\times\dfrac{11}{12}+\dfrac{3}{5}\)
\(=\dfrac{2}{15}+\dfrac{3}{5}\)
\(=\dfrac{11}{15}\)
__________
\(\left(1-\dfrac{3}{4}\right)\times\left(1+\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{4}\times\dfrac{4}{3}\times\dfrac{2}{3}\)
\(=\dfrac{4\times2}{4\times3\times3}\)
\(=\dfrac{2}{3\times3}\)
\(=\dfrac{2}{9}\)
16 tháng 10 2020
a) \(1+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(=\frac{16}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\)
\(=\frac{23}{16}\)
b) \(2-\frac{1}{8}-\frac{1}{12}-\frac{1}{16}\)
\(=\frac{96}{48}-\frac{6}{48}-\frac{4}{48}-\frac{3}{48}\)
\(=\frac{83}{48}\)
c) \(\frac{4}{99}\cdot\frac{18}{5}\div\frac{12}{11}+\frac{3}{5}\)
\(=\frac{4\cdot18\cdot11}{99\cdot5\cdot12}+\frac{3}{5}\)
\(=\frac{4\cdot9\cdot2\cdot11}{9\cdot11\cdot5\cdot4\cdot3}+\frac{3\cdot3}{3\cdot5}\)
\(=\frac{2}{15}+\frac{9}{15}=\frac{11}{15}\)
d) \(\left(1-\frac{3}{4}\right)\left(1+\frac{1}{3}\right)\div\left(1-\frac{1}{3}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{3}\div\frac{2}{3}\)
\(=\frac{1\cdot4\cdot3}{4\cdot3\cdot2}=\frac{1}{2}\)
8 tháng 4 2023
1)\(\dfrac{2}{9}+\dfrac{-3}{4}+\dfrac{5}{30}\)
\(=\dfrac{2.20}{9.20}+\dfrac{-3.45}{4.45}+\dfrac{5.6}{30.6}\)
\(=\dfrac{40}{180}+\dfrac{-135}{180}+\dfrac{30}{180}\)
\(=\dfrac{40+\left(-135\right)+30}{180}\)
\(=\dfrac{-65}{180}\)
\(=\dfrac{-13}{36}\)
2)\(\dfrac{-7}{12}-\dfrac{11}{18}\)
\(=\dfrac{-7.3}{12.3}-\dfrac{11.2}{18.2}\)
\(=\dfrac{-21}{36}-\dfrac{22}{36}\)
\(=\dfrac{-21-22}{36}\)
\(=\dfrac{-43}{36}\)
3)\(\dfrac{7}{8}-\dfrac{-5}{16}\)
\(=\dfrac{7.2}{8.2}-\dfrac{-5}{16}\)
\(=\dfrac{14}{16}-\dfrac{-5}{16}\)
\(=\dfrac{14-\left(-5\right)}{16}\)
\(=\dfrac{19}{16}\)
4)\(\dfrac{3}{8}-\dfrac{-9}{10}-\dfrac{5}{16}\)
\(=\dfrac{3.10}{8.10}-\dfrac{-9.8}{10.8}-\dfrac{5.5}{16.5}\)
\(=\dfrac{30}{80}-\dfrac{-72}{80}-\dfrac{25}{80}\)
\(=\dfrac{30-\left(-72\right)-25}{80}\)
\(=\dfrac{77}{80}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 3 2023
\(\dfrac{7}{6}\) + \(\dfrac{5}{12}\) - \(\dfrac{1}{18}\) - 1
= \(\dfrac{42}{36}\) + \(\dfrac{15}{36}\) - \(\dfrac{2}{36}\) - \(\dfrac{36}{36}\)
= \(\dfrac{19}{36}\)
\(\dfrac{13}{6}\) + \(\dfrac{5}{18}+3-\dfrac{7}{12}\)
= \(\dfrac{78}{36}+\dfrac{10}{36}+\dfrac{108}{36}-\dfrac{21}{36}\)
= \(\dfrac{175}{36}\)
3 + \(\dfrac{11}{4}-\dfrac{1}{12}-\dfrac{3}{16}\)
= \(\dfrac{144}{48}+\dfrac{132}{48}-\dfrac{4}{48}-\dfrac{9}{48}\)
= \(\dfrac{263}{48}\)
\(\dfrac{2}{5}\) x \(\dfrac{7}{4}\) - \(\dfrac{2}{5}\) x \(\dfrac{3}{4}\)
= \(\dfrac{2}{5}\) x ( \(\dfrac{7}{4}\) - \(\dfrac{3}{4}\))
= \(\dfrac{2}{5}\) x \(\dfrac{4}{4}\)
= \(\dfrac{2}{5}\)
27 tháng 7 2021
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
a] 23/16
c)7/45
d)1/2
b)83/48
Đúng nhé mình làm tong toán nâng cao lớp 5 rồi