cho
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\)
\(B=\left(\dfrac{3}{2}\right)^{2013}:2\)
tính B-A
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có : \(\dfrac{3}{2}\)A= \(\dfrac{3}{4}+\)\(\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\)\(...+\left(\dfrac{3}{2}\right)^{2013}\) (1)
A= \(\dfrac{1}{2}+\dfrac{3}{2}\)\(+\left(\dfrac{3}{2}\right)^2+...+\)\(\left(\dfrac{3}{2}\right)^{2012}\) (2)
Lấy (1) trừ đi (2) vế theo vế:
\(\dfrac{3}{2}A-A=\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^{2013}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}-\dfrac{5}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\)
ta có : \(B=\left(\dfrac{3}{2}\right)^{2013}:2=\dfrac{3^{2013}}{2^{2013}}.\dfrac{1}{2}=\dfrac{3^{2013}}{2^{2014}}\)
Vậy \(A-B=\dfrac{3^{2013}}{2^{2014}}-\left(\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\right)\)
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
a) 2021 - (1/3)² . 3²
= 2021 - 1/9 . 9
= 2021 - 1
= 2020
b) 5/10 + 9 . (-3/2)
= 1/2 - 27/2
= -26/2
= -13
c) -10 . (-2021/2022)⁰ + (2/5)² : 2
= -10 . 1 + 4/25 . 2
= -10 + 8/25
= -68/7
\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
a) = 0,16 - 0,064 . (-3)
= 0,16 + 0,192
= 0,352
b) = (7/4)³ - (7/4)² + 1
= 343/64 - 49/16 + 1
= 147/64 + 1
= 211/64
c) = 8/27 - 4.(-7/4)² - 8/27
= -4.49/16
= -49/4
Lời giải:
a. $=0,16-(-0,064).(-3)=0,16-0,192=-0,032$
b. $=(1\frac{3}{4})^2(1\frac{3}{4}-1)+1=(1\frac{3}{4})^2.\frac{3}{4}+1$
$=\frac{147}{64}+1=\frac{211}{64}$
c.
$=(\frac{2}{3})^3-4(\frac{-7}{4})^2-(\frac{2}{3})^3$
$=-4(\frac{-7}{4})^2=\frac{-49}{4}$
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
Ta có \(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\left(1\right)\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{4}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2013}\left(2\right)\)
Lấy (2) - (1) ta được:
\(\dfrac{3}{2}A-A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{1}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}+\dfrac{1}{2}\)
Vậy \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)