`x/2=2/x`

`=>x.x=2.2`

`=>x^2=4`

`=>x=+-2`

\(\left(-12\right).65+7.12-8.\left(-12\right)\)

\(=-12\left(65-7-8\right)\)

\(=-12.50\)

\(=-600\)

\(2^{x+3}.2=2^2.3+52\)

\(=>2^{x+3}.2=64\)

\(=>2^{x+3}=64:2\)

\(=>2^{x+3}=32\)

\(=>2^{x+3}=2^5\)

=>x+3=5

=>x=5-3

=>x=2

Vậy ...........

 2^{x + 3} . 2 = 2² . 3 + 52

 2^{x + 3} . 2 = 4 . 3 + 52

 2^{x + 3} . 2 = 12 + 52

 2^{x + 3} . 2 = 64

     2^{x + 3}  = 64 : 2

     2^{x + 3}  = 32

     2^{x + 3} = 2⁵

     x + 3 = 5

           x = 5 - 3

           x = 2

Vậy x = 2

`-2/4 = 1/(x-1)`

`=> -2(x-1)=4.1`

`=> -2(x-1)=4`

`=> x-1=4:(-2)`

`=> x-1=-2`

`=>x=-2+1`

`=>x=-1`

(n+3) ⋮ (2n-1)

=> 2.(n+3)⋮2n-1

=> 2n+6 ⋮ 2n-1

=> (2n-1)+7⋮2n-1

mà 2n-1⋮2n-1

=> 7⋮2n-1

=>2n-1∈Ư(7)={1;7}

=>2n∈{2;8}

=>n∈{1;4}

Vậy n∈{1;4}

(4n+7)⋮(4n+1)

Vì 4n+7=(4n+1)+6

=>(4n+1)+6⋮(4n+1) mà 4n+1⋮4n+1

=>6⋮4n+1

=>4n+1∈Ư(6)={1;2;3;6}

=>4n∈{0;1;2;5}

=>n∈{0;0,25;0,5;1,25}

Vì n là stn nên n=0

\(=\dfrac{1}{3}:\dfrac{7\cdot\left(5+3\right)}{18\cdot\left(53-11\right)}\) 

\(=\dfrac{1}{3}:\dfrac{7\cdot8}{18\cdot42}\) 

\(=\dfrac{1}{3}:\dfrac{7\cdot2\cdot2\cdot2}{3\cdot3\cdot2\cdot2\cdot3\cdot7}\)

\(=\dfrac{1}{3}:\dfrac{2}{3\cdot3\cdot3}\)

\(=\dfrac{1}{3}\cdot\dfrac{27}{2}\)

\(=\dfrac{9}{2}\)

ĐKXĐ: x^2+30<>0

=>\(x\in R\)

\(6^{2.x+5}=216\)

\(=>6^{2.x+5}=6^3\)

=>2.x+5=3

=>2.x=3-5

=>2.x=-2

=>x=-2:2

=>x=-1

`6^(2x+5)=216`

`=> 6^(2x+5)=6^3`

`=>2x+5=3`

`=>2x=3-5`

`=>2x=-2`

`=>x=-2:2`

`=>x=-1`

`6^(2x+5)=216`

`6^(2x+5)=6^3`

`2x+5=3`

`2x=3-5`

`2x=-2`

`x=-2:2`

`x=-1`