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Lời giải:
\(=-5^{22}-(-222-(-122-100+5^{22}+2022))\)
\(=-5^{22}-(-222+122+100-5^{22}-2022)\)
\(=-5^{22}+222-122-100+5^{22}+2022\)
\(=(-5^{22}+5^{22})+222-(122+100)+2022=0+222-222+2022=2022\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=-5^{22}\left\{-222\left[-122-\left(100-5^{22}\right)+2022\right]\right\}\)
\(A=-5^{22}\left\{-222\left[1900-\left(100-5^{22}\right)\right]\right\}\)
\(A=-5^{22}\left[-222\left(1900-100+5^{22}\right)\right]\)
\(A=-5^{22}\left[-222\left(1800+5^{22}\right)\right]\)
\(A=-5^{22}\left(-399600-222\cdot5^{22}\right)\)
\(A=399600\cdot5^{22}+222\cdot5^{44}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(=-5^{22}+222+[-122-(100-5^{22})+2022]\)
\(=-5^{22}+222-122-100+5^{22}+2022\\ =(-5^{22}+5^{22})+(222-122-100)+2022\\ =0+0+2022=2022\)
Bài 2:
$2n^2+n-6\vdots 2n+1$
$\Rightarrow n(2n+1)-6\vdots 2n+1$
$\Rightarrow 6\vdots 2n+1$
$\Rightarrow 2n+1\in Ư(6)$
Mà $2n+1$ lẻ nên $2n+1\in \left\{\pm 1; \pm 3\right\}$
$\Rightarrow n\in \left\{0; -1; 1; -2\right\}$
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Lời giải:
$=5^{22}-22+[122-(100+5^{22})+2022]$
$=5^{22}-22+122-100-5^{22}+2022$
$=(5^{22}-5^{22})+(-22+122-100)+2022$
$=0+0+2022=2022$
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
215 + 43 - 15 - 23
= (215 - 15) + (43 - 23)
= 200 + 20
= 220
2022 - (122 + 2022) + (122 - 325)
= 2022 - 122 - 2022 + 122 - 325
= (2022 - 2022) - (122 - 122) - 325
= 0 - 0 - 325
= - 325
12.(-137) + 12.136
= 12.( -137 + 136)
= 12.(-1)
= - 12
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a)
\(\begin{array}{l}\left( {13x{\rm{ }}-{\rm{ }}{{12}^2}} \right):{\rm{ }}5{\rm{ }} = {\rm{ }}5\\13x{\rm{ }}-{\rm{ }}{12^2} = 5.5\\13x{\rm{ }}-{\rm{ }}144 = 25\\13x = 25 + 144\\13x = 169\\x = 13\end{array}\)
Vậy \(x = 13\)
b)
\(\begin{array}{l}3x\left[ {{8^2} - 2.\left( {{2^5} - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.\left( {32 - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.31} \right]{\rm{ }} = {\rm{ }}2022\\3x\left( {64 - 62} \right){\rm{ }} = {\rm{ }}2022\\3x.2 = 2022\\6x = 2022\\x = 337\end{array}\)
Vậy \(x = 337.\)
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