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27 tháng 1 2016
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
VM
22 tháng 9 2019
1)
a) \(-\frac{9}{34}:\frac{17}{4}\)
\(=-\frac{18}{289}.\)
b) \(1\frac{1}{2}.\frac{1}{24}\)
\(=\frac{3}{2}.\frac{1}{24}\)
\(=\frac{1}{16}.\)
c) \(-\frac{5}{2}:\frac{3}{4}\)
\(=-\frac{10}{3}.\)
d) \(4\frac{1}{5}:\left(-2\frac{4}{5}\right)\)
\(=\frac{21}{5}:\left(-\frac{14}{5}\right)\)
\(=-\frac{3}{2}.\)
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28 tháng 5 2022
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
8 tháng 8 2018
a) \(\left(\frac{2}{3}+\frac{1}{5}\right)^2:\left(\frac{2}{5}-\frac{1}{3}\right)\)
\(=\left(\frac{13}{15}^2\right)\cdot15\)
\(=\frac{169\cdot15}{225}\)
\(=\frac{169}{15}\)
8 tháng 8 2018
b)
\(\left(2\frac{1}{3}-1\frac{3}{5}\right)\cdot\left(2\frac{4}{9}:3\frac{1}{2}\right)^2\)
\(=\left(\frac{7}{3}-\frac{8}{5}\right)\cdot\left(\frac{22}{9}\cdot\frac{7}{2}\right)^2\)
\(=\frac{11\cdot5929}{15\cdot81}\)
\(=53,6781893\)
NV
26 tháng 6 2018
\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(b,B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)
\(c,C=4+4^2+4^3+...+4^n\)
\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(d,D=1+5+5^2+...+5^{2000}\)
\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(\Rightarrow D=\frac{5^{2001}-1}{4}\)
TC
21 tháng 3 2021
b)
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2
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 \(S=\frac{3^n-1}{4}+\frac{n}{2}\)
LT
1
15 tháng 7 2015
\(A=\frac{2}{5}+\left(-\frac{4}{3}\right)+\left(-\frac{1}{2}\right)\)
\(A=\frac{12}{30}+\left(-\frac{40}{30}\right)+\left(-\frac{15}{30}\right)\)
\(A=-\frac{43}{30}\)
b) \(B=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\left(\frac{1}{4}+\frac{3}{80}\right)\right]\)
\(B=\frac{1}{3}-\left[\left(-\frac{5}{4}\right)-\frac{23}{80}\right]\)
\(B=\frac{1}{3}+\frac{123}{80}\)
\(B=\frac{449}{240}\)
IM
22 tháng 7 2016
a)
\(5A=5+5^2+.....+5^{101}\)
\(\Rightarrow5A-A=\left(5+5^2+.....+5^{101}\right)-\left(1+5+.....+5^{100}\right)\)
\(\Rightarrow4A=5^{101}-1\)
\(\Rightarrow A=\frac{5^{101}-1}{4}\)
b)
\(2B=1+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2^2}+.....+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
a)\(-\dfrac{2}{5}:-\dfrac{3}{4}\\ =-\dfrac{2}{5}\cdot-\dfrac{4}{3}=\dfrac{8}{15}\)
b)\(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)=\dfrac{21}{5}:\left(-\dfrac{14}{5}\right)=-\dfrac{3}{2}\)