a)(a5)7  = a^{5 x 7} = a^{3 5}    

   b)(a3)4.a⁹ = a^{3 x 4} . a⁹ = a⁷ . a⁹ = a^{9 + 7} = a^{1 6}     

      c)(23)5.(23) ⁴ = 2^{3 x 5} . a^{3 x 4} = 2 ¹⁵ . a^{1 2} = 2^{1 5 + 1 2} = 2^{2 7} 

a

=6+12-3.3

=6+12-9

=18+9

=27

27

1/2×X+1/3×X+X=2022

= 1/2×X+1/3×X+Xx1 =2022

= X x ( 1/2 + 1/3 + 1) = 2022

= X x 11/6 = 2022

= X = 2022: 11/6

= X = 12132/11

\(\dfrac{1}{2}\times x+\dfrac{1}{3}\times x+x=2022\\ \Rightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+1\right)\times x=2022\\\Rightarrow \left(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{6}{6}\right)\times x=2022\\ \Rightarrow\dfrac{11}{6}\times x=2022\\ \Rightarrow x=2022:\dfrac{11}{6}\\ \Rightarrow x=2022\times\dfrac{6}{11}\\ \Rightarrow x=\dfrac{12132}{11}\)

 x × 1/3-1/4×x=1/2

= X x ( 1/3 - 1/4 ) = 1/2

= X x 1/12 = 1/2

= X = 1/2 : 1/12

= X = 6

1/3-1/4.x=1/2

1/4.x=1/3-1/2

1/4.x=-1/6

x=-1/6:1/4

x=-2/3

Lời giải:

$A=1+3+3^2+3^3+...+3^{2022}$

$=1+(3+3^2)+(3^3+3^4)+...+(3^{2021}+3^{2022})$

$=1+3(1+3)+3^3(1+3)+....+3^{2021}(1+3)$

$=1+4(3+3^3+...+3^{2021})$

$=1+4[3+(3^3+3^5)+(3^7+3^9)+....+(3^{2019}+3^{2021})]$

$=13+4[3^3(1+3^2)+3^7(1+3^2)+...+3^{2019}(1+3^2)]$
$=13+40(3^3+3^7+...+3^{2019})$
$=13+40[3^3+(3^7+3^{11})+(3^{15}+3^{19})+...+(3^{2015}+3^{2019})]$
$=1093+40[3^7(1+3^4)+3^{15}(1+3^4)+....+3^{2015}(1+3^4)]$

$=1093+40.82(3^7+3^{15}+...+3^{2015})$
$=1093+16.5.41(3^7+...+3^{2015})$

$=5+16.68+16.5.41(3^7+...+3^{2015})$

Vậy biểu thức chia $16$ dư $5$

\(x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{6}=-1\\ =>\dfrac{6x}{6}+\dfrac{3x}{6}+\dfrac{2x}{6}+\dfrac{x}{6}=-1\\ =>\dfrac{6x+3x+2x+x}{6}=-1\\ =>\dfrac{12x}{6}=-1\\ =>2x=-1\\ =>x=-\dfrac{1}{2}\)

\(x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{6}=-1\)

\(\dfrac{6x+3x+2x+x}{6}=-1\)

\(\dfrac{12x}{6}=-1\)

\(2x=-1\)

\(x=-\dfrac{1}{2}\)

x-135:45=2006

x-3=2006

x=2006+3

x=2009

`x-135:45=2006`

`=>x-3=2006`

`=>x=2006+3=2009`

\(\dfrac{m}{2}-\dfrac{2}{n}=\dfrac{1}{2}\left(n\ne0\right)\\ =>\dfrac{mn-4}{2n}=\dfrac{n}{2n}\\ =>mn-4=n\\ =>n\left(m-1\right)=4\)

\(\begin{matrix}n&1&4&-1&-4\\m-1&4&1&-4&-1\\m&5&2&-3&0\end{matrix}\)

Vậy \(\left(m;n\right)=\left(5;1\right);\left(2;4\right);\left(-3;-1\right);\left(0;-4\right)\)

\(\begin{matrix}n&2\\m-1&2\\m&3\end{matrix}\)