A = \(\frac{10}{\left(1+i\right)^1}\)+ \(\frac{10}{\left(1+i\right)^2}\)+ \(\frac{10}{\left(1+i\right)^3}\)+ \(\frac{10}{\left(1+i\right)^4}\)+ \(\frac{10}{\left(1+i\right)^5}\)= ?
(phương pháp tính ) i = ?
K
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28 tháng 6 2016
a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
28 tháng 6 2016
c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)
d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)
19 tháng 3 2020
a) \(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
= \(\frac{-1}{24}-\left[\frac{6}{24}-\left(\frac{12}{24}-\frac{21}{24}\right)\right]\)
= \(\frac{-1}{24}-\left[\frac{6}{24}-\frac{-9}{24}\right]\)
= \(\frac{-1}{24}-\frac{15}{24}\)
= \(\frac{-16}{24}\) = \(\frac{-2}{3}\)
19 tháng 3 2020
b) \(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(-\frac{2}{7}-\frac{1}{10}\right)\right]\)
= \(\left(\frac{50}{70}-\frac{98}{70}\right)-\left[\frac{35}{70}-\left(-\frac{20}{70}-\frac{7}{70}\right)\right]\)
= \(\frac{-48}{70}-\left[\frac{35}{70}-\left(-\frac{20}{70}-\frac{7}{70}\right)\right]\)
= \(\frac{-48}{70}-\left[\frac{35}{70}-\frac{-27}{70}\right]\)
= \(\frac{-48}{70}-\frac{62}{70}\)
= \(\frac{-110}{70}=\frac{-11}{7}\)
28 tháng 6 2022
Bài 1:
\(\Leftrightarrow-\dfrac{5}{7}:x=-\dfrac{7}{18}-\dfrac{1}{6}=\dfrac{-7}{18}-\dfrac{3}{18}=\dfrac{-10}{18}=\dfrac{-5}{9}\)
=>x=5/9:5/7=7/9
Bài 2:
a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}=\dfrac{1000}{2}=500\)
b: \(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\)
\(=-\dfrac{1}{1000}\)
AH
Akai Haruma
Giáo viên
1 tháng 1 2020
Bài 1:
a)
\((\frac{3}{5})^2-[\frac{1}{3}:3-\sqrt{16}.(\frac{1}{2})^2]-(10.12-2014)^0\)
\(=\frac{9}{25}-(\frac{1}{9}-1)-1\)
\(=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\)
b)
\(|-\frac{100}{123}|:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})\)
\(=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}=(\frac{100}{123}+\frac{23}{123}):\frac{4}{3}=1:\frac{4}{3}=\frac{3}{4}\)
c)
\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)
\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)
AH
Akai Haruma
Giáo viên
1 tháng 1 2020
Bài 2:
a)
\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)
b)
\((\frac{1}{2}-x)^2=(-2)^2=2^2\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)
c)
\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)
\(|3x+\frac{1}{2}|=\frac{2}{3}+1=\frac{5}{3}\)
\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)
d)
\(3^{2x+1}=81=3^4\)
\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)
14 tháng 10 2016
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
VD
13 tháng 10 2016
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
NN
17 tháng 12 2015
a) \(=10\frac{1}{4}\cdot\frac{-5}{3}-8\frac{1}{4}\cdot\frac{-5}{3}-5=\left(10\frac{1}{4}-8\frac{1}{4}\right)\cdot\frac{-5}{3}-5\)
\(=\left(\frac{41}{4}-\frac{33}{4}\right)\cdot\frac{-5}{3}-5=2\cdot\frac{-5}{3}-5\)\(=\frac{-10}{3}-\frac{15}{3}=\frac{-25}{3}\)
b)\(=\frac{5}{7}+1+\frac{2}{7}+\frac{2^{10}\cdot\left(2^3\right)^3}{\left(2^2\right)^9}\)
\(=\frac{5}{7}+\frac{2}{7}+1+\frac{2^{10}\cdot2^9}{2^{27}}\)
\(=1+1+\frac{1}{2^8}=2+\frac{1}{256}=\frac{512}{256}+\frac{1}{256}=\frac{513}{256}\)