2:2+1+30:2+9-6+52:9-4+9-20+999-345+335-513:20+639-256
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Những câu hỏi liên quan
TG
2 tháng 8 2020
1) \(373737.73-37.737373\)
\(=37.10101.73-37.73.10101\)
\(=0\)
2) \(19.41414141-41.19191919\)
\(=19.41.1010101-41.19.1010101\)
\(=0\)
3) \(123456.456789456789-456789.123456123456\)
\(=123456.456789.1000001-456789.123456.1000001\)
\(=0\)
4) \(250.12.7\)
\(=250.4.3.7\)
\(=\left(250.4\right).\left(3.7\right)\)
\(=1000.21\)
\(=21000\)
5) \(24.125.9\)
\(=3.8.125.9\)
\(=\left(3.9\right).\left(8.125\right)\)
\(=27.1000\)
\(=27000\)
6) \(125.44.18\)
\(=125.4.11.2.9\)
\(=\left(125.4.2\right).\left(11.9\right)\)
\(=1000.99\)
\(=99000\)
7) \(8.24.6+12.4.52+2.22.24\)
\(=48.24+48.52+48.22\)
\(=48\left(24+52+22\right)\)
\(=48.98\)
\(=48\left(100-2\right)\)
\(=48.100-48.2\)
\(=4800-96\)
\(=4704\)
8) \(18.58.2-9.27.4+12.69.3\)
\(=36.58-36.27+36.69\)
\(=36\left(58-27+69\right)\)
\(=36.100=3600\)
9) \(30.85.3+18.72.5-45.37.2-9.20.10\)
\(=90.85+90.72-90.37-90.20\)
\(=90\left(85+72-37-20\right)\)
\(=90.100=9000\)
1 tháng 10 2021
\(a,=75+60-13=122\\ b,=250-25+40=265\\ c,=4+100-9=95\\ d,=45+80-27=98\\ e,=5^3-100=125-100=25\\ g,=5+5=10\)
C
22 tháng 12 2023
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
17 tháng 2 2020
\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)
\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)
vậy M=20
ND
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 10 2023
\(A=1+\dfrac{1}{2}+1+\dfrac{1}{4}+1+\dfrac{1}{8}+...+1+\dfrac{1}{256}+1+\dfrac{1}{512}=\)
\(=1x\left(\dfrac{512-2}{2}+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{...1}{256}+\dfrac{1}{512}\right)=\)
\(256+B\)
\(2B=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(B=2B-B=1+\dfrac{1}{1024}\)
\(\Rightarrow A=265+1+\dfrac{1}{1024}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
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