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\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
\(\frac{2x+1}{3}=\frac{5}{2}\)
\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)
2x= 15/2 - 1 = 13/2
x = 13/2 : 2
x = 13/4
b) 2x + 2x+1 + 2x+2 + 2x+3 = 480
2x.(1+ 2 +22 + 23) = 480
2x . 15 = 480
2x = 480 : 15 = 32
2x = 25 => x = 5
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)
< = > 3x= -6 => x = -2
a) \(\frac{x-2}{3}=\frac{x+1}{4}\)
=> (x - 2).4 = 3.(x + 1)
=> 4x - 8 = 3x + 3
=> 4x - 3x = 3 + 8
=> x = 11
Vậy x = 11
b) \(2.\left(x+3\right)-\frac{1}{2}=x-1\)
=> \(2x+6-\frac{1}{2}=x-1\)
=> \(2x+\frac{11}{2}=x-1\)
=> \(2x-x=-1-\frac{11}{2}\)
=> \(x=-\frac{13}{2}\)
Vậy \(x=-\frac{13}{2}\)
Ta có: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
\(Q=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+10}\)
\(Q=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+....+\frac{1}{\frac{10.\left(10+1\right)}{2}}\)
\(Q=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+....+\frac{1}{\frac{10.11}{2}}\)
\(Q=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)
\(\frac{1}{2}Q=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
\(\frac{1}{2}Q=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
=>\(Q=\frac{9}{22}.2=\frac{9}{11}\)
\(Q=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{55}\\ \Rightarrow\frac{1}{2}Q=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
Tiếp theo tự tính nhé
theo phân số: 55/21;134/43;116/37;77/21
theo số thập phân:2.(619047); 3.(11627...); 3.(135); 4.(0526...)
Like nha !
| x | 7 | 9 | |||
| x2 | 49 | 81 | |||
| x2-49 | - | 0 | + | + | + |
| x2-81 | - | - | - | 0 | + |
| A | + | 0 | - | 0 | + |
dựa vào bảng ta có khi 7<x<9 thì A<0 vậy 7<x<9
b, ta có : \(\frac{2015}{1}\)+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+......+\(\frac{1}{2015}\)
=1+1+1+1......+1+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+.......+\(\frac{1}{2015}\)
(2015 số 1)
=1+(1+\(\frac{2014}{2}\))+(1+\(\frac{2013}{3}\))+........+(1+\(\frac{1}{2015}\))
=\(\frac{2016}{2016}\)+\(\frac{2016}{2}\)+\(\frac{2016}{3}\)+.........+\(\frac{2016}{2015}\)
=2016(\(\frac{1}{2016}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.........+\(\frac{1}{2015}\))
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)=1-2^{2016}< 1\)
=>đpcm
2. \(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.\left(1993+1\right)-1}{1993.1995+1994}=\frac{1995.1993+1995-1}{1993.1995+1994}=\frac{1995.1993+1994}{1993.1995+1994}\)
1. \(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}\)
\(=\frac{7-3}{3.7}+\frac{12-7}{7.12}+\frac{13-12}{12.13}+\frac{23-20}{20.23}\)
\(=\left[\frac{7}{3.7}-\frac{3}{3.7}\right]+\left[\frac{12}{7.12}-\frac{7}{7.12}\right]+\left[\frac{13}{12.13}-\frac{12}{12.13}\right]+\left[\frac{20}{13.20}-\frac{13}{13.20}\right]+\left[\frac{23}{20.23}-\frac{20}{20.23}\right]\) \(=\left[\frac{1}{3}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{12}\right]+\left[\frac{1}{12}-\frac{1}{13}\right]+\left[\frac{1}{13}-\frac{1}{20}\right]+\left[\frac{1}{20}-\frac{1}{23}\right]\) \(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\) \(=\frac{1}{3}-\frac{1}{23}\\ =\frac{20}{69}\)