(1/2+2/3 -3/4 ):1/24
K
Khách
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TP
2
27 tháng 8 2023
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):\left(\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}\right)+\left(\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}\right):\left(\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}\right)\)
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right)\)
\(D=\dfrac{1}{3}+\dfrac{2}{3}\)
\(D=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 8 2023
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}}\) + \(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}}{\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}}\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3.\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}\) + \(\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}\)
D = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)
D = \(\dfrac{3}{3}\)
D = 1
10 tháng 12 2023
1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
S
4 tháng 3 2023
\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{4}{12}+\dfrac{9}{12}+\dfrac{6}{12}=\dfrac{19}{12}\)
\(\dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{3}{4}+\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{5}{6}=\dfrac{10}{6}=\dfrac{5}{3}\)
\(\dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{6}{10}+\dfrac{15}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)
VT
4 tháng 3 2023
\(\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{13}{12}+\dfrac{1}{2}=\dfrac{19}{12}\\ \dfrac{6}{8}+\dfrac{2}{4}+\dfrac{6}{24}=\dfrac{5}{4}+\dfrac{6}{24}=\dfrac{3}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{5}{6}+\dfrac{5}{6}=\dfrac{10}{6}\\ \dfrac{3}{5}+\dfrac{3}{2}+2=\dfrac{21}{10}+\dfrac{20}{10}=\dfrac{41}{10}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 3 2020
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
LH
4
KN
21 tháng 8 2020
1. \(\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^4=\left(-\frac{1}{2}\right)^5=-\frac{1}{32}\)
2. \(6.3^2-24:2^3=6.9-24:8=54-3=51\)
3. \(\left(\frac{1}{4}\right)^2.\left(\frac{1}{4}\right)^3=\left(\frac{1}{4}\right)^5=\frac{1}{1024}\)
21 tháng 8 2020
1) (-1/2).(-1/2)^4 = ( -1/2)^ 5 = -1/32
2) 6.3^2 - 24:2^3 = 6,9 - 24 : 8 = 54 - 3 = 51
3) (1/4)^2 . (1/4)^3 = ( 1/4)^5 = 1/1024
9 tháng 1 2018
1, 54 : x - 1 = 5
54 : x = 5+1 = 6
x = 54 : 6 = 9
2, 42 : x + 0 = 8
x = 42 : 8 = 21/4
3, 24 : x - 8 = 0
24 : x = 0 + 8 = 8
x = 24 : 8 = 3
Tk mk nha
7 tháng 1 2017
\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)
\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)
\(=\left(\frac{6}{12}+\frac{8}{12}-\frac{9}{12}\right).24\)
\(=\frac{6+8-9}{12}.24=\frac{5}{12}.24=10\)
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