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1) Ta có
\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)
\(C=\dfrac{1}{2022}\)
2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)
A = - 522 - { - 222 - [ - 122 - (100 - 522) + 2022] }
A = - 522 - { -222 - [- 122 - 100 + 522 ] + 2022}
A = - 522 - { -222 - { - 222 + 522 } + 2022}
A = - 522 - {- 222 + 222 - 522 + 2022}
A = -522 + 522 - 2022
A = - 2022
B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)
B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2
B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2
B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)
B = \(\dfrac{2+3+4+...+21}{2}\)
B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)
B = \(\dfrac{23\times20:2}{2}\)
B = \(\dfrac{23\times10}{2}\)
B = 23
\(\dfrac{1}{2}A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}\)
\(A-\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(A=\dfrac{1}{2^{2022}}-2\)
12A=12+(12)2+(12)3+(12)4+...+(12)202312A=12+(12)2+(12)3+(12)4+...+(12)2023
A−12A=(12)2023−1A−12A=(12)2023−1
12A=(12)2023−112A=(12)2023−1
A=122022−2
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: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
`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`
\(A=\dfrac{-5}{3}\cdot\dfrac{11}{2}\cdot\dfrac{4}{3}=\dfrac{-20\cdot11}{2\cdot9}=\dfrac{-110}{9}\)
\(B=\dfrac{2}{4}\left(\dfrac{4}{11\cdot15}+\dfrac{4}{15\cdot19}+...+\dfrac{4}{51\cdot55}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+...+\dfrac{1}{51}-\dfrac{1}{55}\right)\)
=1/2*4/55
=2/55
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
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 \(A-B=\dfrac{3^{2013}}{2^{2014}}-\left(\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\right)\)
Lời giải:
$A-\frac{1}{2}=\frac{3}{2}+(\frac{3}{2})^2+...+(\frac{3}{2})^{2022}$
$\frac{3}{2}(A-\frac{1}{2})=(\frac{3}{2})^2+(\frac{3}{2})^3+...+(\frac{3}{2})^{2023}$
$\Rightarrow \frac{3}{2}(A-\frac{1}{2})-(A-\frac{1}{2})=(\frac{3}{2})^{2023}-\frac{3}{2}$
$\Rightarrow \frac{1}{2}(A-\frac{1}{2})=(\frac{3}{2})^{2023}-\frac{3}{2}$
$\Rightarrow \frac{1}{2}A=(\frac{3}{2})^{2023}-\frac{1}{2}$
$\Rightarrow A=2(\frac{3}{2})^{2023}-1$