Lời giải:
\(=-5^{22}-(-222-(-122-100+5^{22}+2022))\)

\(=-5^{22}-(-222+122+100-5^{22}-2022)\)

\(=-5^{22}+222-122-100+5^{22}+2022\)

\(=(-5^{22}+5^{22})+222-(122+100)+2022=0+222-222+2022=2022\)

\(A=-5^{22}\left\{-222\left[-122-\left(100-5^{22}\right)+2022\right]\right\}\)

\(A=-5^{22}\left\{-222\left[1900-\left(100-5^{22}\right)\right]\right\}\)

\(A=-5^{22}\left[-222\left(1900-100+5^{22}\right)\right]\)

\(A=-5^{22}\left[-222\left(1800+5^{22}\right)\right]\)

\(A=-5^{22}\left(-399600-222\cdot5^{22}\right)\)

\(A=399600\cdot5^{22}+222\cdot5^{44}\)

Lời giải:

$=5^{22}-22+[122-(100+5^{22})+2022]$

$=5^{22}-22+122-100-5^{22}+2022$

$=(5^{22}-5^{22})+(-22+122-100)+2022$

$=0+0+2022=2022$

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 \(x = 337.\)

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 $2n+1\in \left\{\pm 1; \pm 3\right\}$

$\Rightarrow n\in \left\{0; -1; 1; -2\right\}$

\(50\%+\dfrac{7}{12}-\dfrac{1}{2}\)

\(=\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\)

\(=\dfrac{7}{12}\)

_______________

\(2022\cdot67+2022\cdot43-2022\cdot10\)

\(=2022\cdot\left(67+43-10\right)\)

\(=2022\cdot100\)

\(=202200\)

_____________________

\(10,3+6,9+8,7+13,1\)

\(=\left(13,1+6,9\right)+\left(10,3+8,7\right)\)

\(=20+19\)

\(=39\)

___________________

\(17,58\times43+57\times17,58\)

\(=17,58\times\left(43+57\right)\)

\(=17,58\times100\)

\(=1758\)

tớ ghi sai nên cập nhật lại câu hỏi, phần 3 các bạn trả lời sớm mà bị sai thì bỏ qua cho tớ nhé

bạn ơi , bạn có nhầm ko , kết quả sơ sơ thôi cũng lớn hơn số này r

mk thấy đề hơi sai sai bạn à

 a)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)

b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)

c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)

\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)

\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)

Tương tự câu d,e,f bạn tự làm nhé

a: \(12+2^2+3^2+4^2+5^2\)

\(=12+4+9+16+25\)

\(=16+50=66\)

\(\left(1+2+3+4+5\right)^2=15^2=225\)

=>\(12+2^2+3^2+4^2+5^2< \left(1+2+3+4+5\right)^2\)

b: \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2< \left(1+2+3+4\right)^3\)

c: \(5^{202}=5^2\cdot5^{200}=25\cdot5^{200}>16\cdot5^{200}\)

d: \(18\cdot4^{500}=18\cdot2^{1000}\)

\(2^{1004}=2^4\cdot2^{1000}=16\cdot2^{1000}\)

=>\(18\cdot4^{500}>2^{1004}\)

e: \(2022\cdot2023^{2024}+2023^{2024}=2023^{2024}\left(2022+1\right)\)

\(=2023^{2025}\)

dạ em nhầm ạ phần a) số 12 phải là 1²