16 - 10 + 9 + 1 = ???
A. 10
B. 20
C. 30
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P
2
8 tháng 4 2023
a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{2}{3}\)
b: \(=\dfrac{11^9\cdot3^{18}}{3^{18}\cdot11^{11}}=\dfrac{1}{121}\)
P
1
5 tháng 4 2023
a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{20}{30}=\dfrac{2}{3}\)
b: \(=\dfrac{11^9\cdot3^{18}}{3^{10}\cdot11^{11}\cdot3^8}=\dfrac{1}{121}\)
N
3
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
a) Ta có: \(10^{20}=\left(10^2\right)^{10}=100^{10}\)
Mà \(100^{10}>19^{10}\)
\(\Rightarrow10^{20}>19^{10}\)
b) Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
Mà: \(125^{10}< 243^{10}\)
\(\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) Ta có: \(64^8=\left(2^6\right)^8=2^{48}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Mà: \(2^{48}=2^{48}\)
\(\Rightarrow64^8=16^{12}\)
3 tháng 7 2023
a) 1020và 1910
Ta có: 1020= (102)10 và 1910
= 10010 và 1910
Vì 10010>1910 => 1020>1910
b) (-5)30 và (-3)50
Ta có:
(-5)30= [(-5)3]10=(-125)10 và (-3)50=[(-3)5]10=(-243)10
Vì -12510>-24310 Nên (-5)30>(-3)50
c) 648 và 1612
= (43)8và (42)12
= 424 và 424
=> 648 = 1612
TT
1
1 tháng 3 2022
a: =5/4-3/4=2/4=1/2
b: =30/45-18/45=12/45=4/15
c: =10/12-9/12=1/12
d: =4/3-1/4=16/12-3/12=13/12
NA
Ngoc Anh Thai
Giáo viên
28 tháng 3 2021
a)
\(\dfrac{2}{3}x-\dfrac{1}{5}x=-\dfrac{7}{10}\\ \left(\dfrac{2}{3}-\dfrac{1}{5}\right)x=-\dfrac{7}{10}\\ \dfrac{7}{15}x=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}:\dfrac{7}{15}\\ x=-\dfrac{15}{10}\\ x=-\dfrac{3}{2}.\)
b)
\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right).\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{2021}-1\right)\\ =\dfrac{-9}{10}.\dfrac{-10}{11}.\dfrac{-11}{12}...\dfrac{-2020}{2021}\\ =\dfrac{9}{2021}.\)
c)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\\ =\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ =\dfrac{1}{3}-\dfrac{1}{10}\\ =\dfrac{7}{30}.\)
(câu c em ghi nhầm đề 1 chút nhé, chỗ đó là 56 chứ không phải 50)
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
H9
HT.Phong (9A5)
CTVHS
9 tháng 8 2023
a) \(\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{10}{16}+\dfrac{10}{24}\)
\(=\dfrac{3}{8}+\dfrac{7}{12}+\dfrac{5}{8}+\dfrac{5}{12}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{7}{12}+\dfrac{5}{12}\right)\)
\(=1+1\)
\(=2\)
b) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{14}{6}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}+\dfrac{19}{13}+\dfrac{1}{9}+\dfrac{7}{3}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}+\dfrac{1}{9}\right)\)
\(=3+2+2\)
\(=7\)
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{7}\)
\(=\dfrac{6}{7}\)
16
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