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27 tháng 1 2016
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
26 tháng 6 2023
15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9
Sửa đề:
A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007
=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007
=-1+1+1/2007=1/2007
16:
\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)
=1/64
17:
=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7
=-6/7
4 tháng 4 2020
1; = ( -4/10 + 3/10 ) : ( -2/5 + 2/3 ) = -1/10 : ( -6/15 + 10/15 ) = -1/10 : 4/15 = -1/10 . 15/4 = -15/40 = -3/8
2; = 25/2 . -5/7 + 39/4 + -3/2 . 5/7 = -125/14 + 39/4 + -15/14 = ( -125/14 + -15/14 ) + 39/4 = -10 + 39/4 = -40/4 + 39/4 = -1/4
3; = 5/52 + 35/52 + 40/52 = 40/52 + 40/52 = 80/52 = 20/13
4; = ( -39/52 + 20/52 ) . 7/2 - ( 117/52 + 32/52 ) . 7/2 = -19/52 . 7/2 - 149/52 . 7/2 = ( -19/52 + -149/52 ) . 7/2 = -168/52 .7/2 = -147/13
5; = ( 36/12 + -9/12 + 8/12 ) - ( -12/6 + -8/6 + -9/6 ) - ( 6/6 - 14/6 - 27/6 ) = 35/12 + 10/12 + 70/12 = 115/12
6; = -1/3 + -8/35 +-2/9 + -1/135 +4/5 +-4/9 +3/7 = (-1/3 + -2/9 + -4/9 ) + ( -8/35 + 4/5 + 3/7 ) + -1/135 = ( -1/3 + -2/3 ) + ( -8/35 + 28/35 + 15/35 ) + -1/135 = -1 + 1 + -1/135 = -1/135
AP
1
5 tháng 7 2023
1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
17 tháng 9 2020
\(\left|\frac{5}{-4}\right|-\left|\frac{1}{-3}\right|+-\frac{5}{6}-4\frac{1}{2}\)
\(=\left|-\frac{5}{4}\right|-\left|\frac{-1}{3}\right|+\frac{-5}{6}-\frac{9}{2}\)
\(=\frac{5}{4}-\frac{1}{3}+\frac{-5}{6}-\frac{9}{2}=-\frac{53}{12}\)
\(\frac{5}{8}-\left|-\frac{1}{12}\right|-3\frac{1}{4}+\left|-\frac{5}{6}\right|\)
\(=\frac{5}{8}-\frac{1}{12}-\frac{13}{4}+\frac{5}{6}=-\frac{15}{8}\)
\(\frac{3}{-7}+\left|-\frac{5}{12}\right|+3\frac{1}{4}+\left|-\frac{5}{6}\right|\)
\(=\frac{-3}{7}+\frac{5}{12}+\frac{13}{4}+\frac{5}{6}=\frac{57}{14}\)
\(1\frac{3}{5}-\left|\frac{1}{-4}\right|+\frac{2}{-3}-\left|-\frac{1}{2}\right|\)
\(=\frac{8}{5}-\left|\frac{-1}{4}\right|+\frac{-2}{3}-\frac{1}{2}\)
\(=\frac{8}{5}-\frac{1}{4}+\frac{-2}{3}-\frac{1}{2}\)
\(=\frac{27}{20}+\frac{-7}{6}=\frac{27}{20}-\frac{7}{6}=\frac{11}{60}\)
PD
0
27 tháng 7 2021
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2018
\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
Trừ theo vế:
\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)
\(4B=5^{2010}-1\)
\(B=\frac{5^{2010}-1}{4}\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2018
\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)
\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)
\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)
Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)
\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)
Trừ theo vế:
\(3X-X=3^n-3^0=3^n-1\)
\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)