n + 15 ⋮ n

⇔ 15 ⋮ n

⇔ n ϵ Ư(15)

15 = 3.5 ⇒ Ư(15) = { 1; 3; 5; 15}

⇔ n ⋮ { 1; 3; 5; 15}

    n + 15 ⋮ n + 2

⇔ n + 2 + 13 ⋮ n + 2

⇔              13 ⋮ n + 2

⇔ n + 2  ϵ { 1; 13}

⇔ n = 11

150:[55-(16-11)²]+2020⁰

=150:[55-5²]+1

=150:[55-25]+1

=150:30+1

=5+1=6

  150 : [ 55 - (16 -11)² ] + 2022⁰

=150 : [ 55- 5² ] + 1

=150 : [ 55-25] +1

= 150 : 30 + 1

= 5 + 1

= 6

Giải 

\(\dfrac{4}{7}x-\dfrac{9}{2}=\dfrac{3}{7}x-\dfrac{5}{2}\\ =>\dfrac{4}{7}x-\dfrac{3}{7}x=-\dfrac{5}{2}+\dfrac{9}{2}\\ =>\left(\dfrac{4}{7}-\dfrac{3}{7}\right)x=\left(\dfrac{-5+9}{2}\right)\\ =>\dfrac{1}{7}x=\dfrac{4}{2}\\ =>\dfrac{1}{7}x=2\\ =>x=2:\dfrac{1}{7}\\ =>x=14\)

Vậy \(x=14\)

\(\dfrac{4}{7}\cdot x-\dfrac{9}{2}=\dfrac{3}{7}\cdot x-\dfrac{5}{2}\)

\(\dfrac{4}{7}\cdot x-\dfrac{3}{7}\cdot x=-\dfrac{5}{2}+\dfrac{9}{2}\)

\(x\left(\dfrac{4}{7}-\dfrac{3}{7}\right)=2\)

\(x\cdot\dfrac{1}{7}=2\)

\(x=2:\dfrac{1}{7}=14\)

Vậy \(x=14\)

$A=1-2+3-4+...+49-50$

$=(1-2)+(3-4)+...+(49-50)$

$=(-1)+(-1)+...+(-1)$

Ta thấy từ $1\to 50$ có $50$ số

$\Rightarrow$ Có $25$ cặp

$\Rightarrow (-1)+(-1)+...+(-1)$

$=(-1).25$

$=-25$

B = \(\dfrac{1}{10.11}+\dfrac{1}{11.12}+......+\dfrac{1}{99.100}\)

B = \(\dfrac{1}{10.}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

B = 1/10 - 1/100

B = 9/100

x ⋮ 18; x ⋮ 40

⇔ x ϵ B(18,40)

18 = 2.3²

40 = 2³ .5

BCNN(18,40 ) = 2³ . 3² . 5 = 360

B(18; 40) ={ 0; 360; 720;....;}

x = 0

\(93x2y\) ⋮2; 5; 9

⇔ \(\left\{{}\begin{matrix}y=0\\9+3+2+x+0⋮9\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}y=0\\14+x⋮9\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)

[2^x + 59]: 3 = 25

2^x + 59      = 25 x 3

2^x + 59     =   75

2^x             = 75 - 59

2^x             =  16

2^x            = 2⁴

 x              = 4