3A=31+32+33+...+398+399+3100+3101
3A-A=3101-1
Vậy 2A=3101-1
3A=31+32+33+...+398+399+3100+3101
3A-A=3101-1
Vậy 2A=3101-1
Q = \(\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{7}{8}+\dfrac{15}{16}+\dfrac{31}{32}+\dfrac{63}{64}+\dfrac{127}{128}-6\)
Giải giùm mấy bài này với ạ
A = \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}-\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
B = \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{9}{8}+\dfrac{17}{16}+\dfrac{33}{32}+\dfrac{65}{64}-7\)
Ai giải nhanh và đúng nhất sẽ được 2 tick cho mỗi comment ạ :33
Tính :
1, \(2016-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}\)
2, \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{9}{8}+\dfrac{17}{16}+\dfrac{33}{32}+\dfrac{65}{64}-7\)
3, A= \(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}-...-\dfrac{1}{9\cdot10}\)
help me ~ mai mik phải nộp rồi
\(\dfrac{2^{18}}{32^3}.\dfrac{27^4}{54^3}\) \(\dfrac{x+1}{9}=\dfrac{4}{x+1}\) Tìm x, y biết:
\(\dfrac{40^4}{120^4}:\dfrac{130^3}{390^3}\) \(19.5^{2x+7}=475\) 5x=4y và y-x=7
\(2^x.8\) \(\dfrac{2x}{5}=\dfrac{10}{x}\)
\((\dfrac{1}{2})^x=\dfrac{1}{32}\) \(33^x:11^x=243\)
CHo A=1/2.2+1/4.4+1/6.6+...+1/2020.2020 CMR A<9/32
1/A=|-3,75|-31/2+1/4
2/A=3/2-(-3)-(2,5-3)-[1,5+(-2,5)]
Tính giá trị biểu thức :
A=\(\left[\dfrac{1\dfrac{11}{31}.4\dfrac{3}{7}-\left(15-6\dfrac{1}{3}.\dfrac{2}{19}\right)}{4\dfrac{5}{6}+\dfrac{1}{6}\left(12-5\dfrac{1}{3}\right)}\right].\dfrac{31}{50}\)
a)1/2x+3/5x=-33/22
b) (2/3x -4/7).(1/2+-3/7:x)=0
c) (4/5-2x).(1/3+3/5:x)=0
Chứng minh rằng a = 5n+2 +5n+1 +5n chia hết cho 31
1)\(\dfrac{2^{18}}{32^3}.\dfrac{27^4}{54^3}\) | 4)\((\dfrac{1}{2})^x=\dfrac{1}{32}\) |
2)\(\dfrac{40^4}{120^4}.\dfrac{130^3}{390^3}\) | 5)\(\dfrac{x+1}{9}=\dfrac{4}{x+1}\) |
3)\(2^x=8\) | 6)\(\dfrac{2x}{5}=\dfrac{10}{x}\) |
7)\(19.5^{2x+7}=475\) | |
8)\(33^x:11^x=243\) | |
9)Tìm x, y biết: 5x=4y và y-x=7 |