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

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

Số thứ nhất `=1+4x1`

Số thứ hai `=1+4x2`

Số thứ ba `=1+4x3`

...

Số thứ `23` `=1+4x23=93`

Số hạng thứ 23 của tổng a là:

     ( 23-1) x 4 + 5 = 93

C

c, 48

`x-135:45=200`

`=>x-3=200`

`=>x=200+3=203`

2006 á

\(\left(x+1\right)^4=\left(2x\right)^4=\left(\pm2x\right)^4\\ =>\left[{}\begin{matrix}x+1=2x\\x+1=-2x\end{matrix}\right.=>\left[{}\begin{matrix}x=1\left(TM\right)\\x=-\dfrac{1}{3}\left(KTM\right)\end{matrix}\right.\)

\(\left(2x-1\right)^5=x^5\\ =>2x-1=x\\ =>2x-x=1\\ =>x=1\left(TM\right)\)