\(=25\times\left(89+10+1\right)\\ =25\times100\\ =2500\)

25x(89+1+10)=25x100=2500

\(n+3⋮n-2\\ =>n-2+5⋮n-2\\ =>n-2\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(=>\left\{{}\begin{matrix}n-2=-1\\n-2=1\\n-2=5\\n-2=-5\end{matrix}\right.=>\left\{{}\begin{matrix}n=1\\n=3\\n=7\\n=-3\end{matrix}\right.\)

\(\dfrac{n+3}{n-2}=1+\dfrac{5}{n-2}\)

Để n+3 chia hết cho n-2 thì 5 phải chia hết cho n- 2. Hay \(n-2=\pm5\) hoặc \(n-2=\pm1\)

Vậy: n =\(\left\{-3;1;3;7\right\}\)

`2^x :2^5=1`

`2^x = 1*2^5`

`2^x=2^5`

`=>x=5`

Vậy `x=5`

2^x : 2⁵ = 1

2^x = 1 . 2⁵

2^x = 2⁵

x=5

3^x-3 - 3² = 2 x 3² 

3^x-3 - 3² = 2 x 9

3^x-3 - 9 = 18

3^x-3 = 18 + 9

3^x-3 = 27

3^x-3 = 3³ 

x-3=3

x=3+3

x=6

3^x-3 = 2. 9 + 9

3^x-3 = 27

3^x-3 = 3³

^{=> x - 3 = 3}

  ^{x = 6}

`28 . x - x . 17 - x = 250`

`x*(28-17-1)=250`

`x* 10=250`

`x =250/10=25`

Vậy `x=25`

= x.(28-17-1)=250

x.10=250

x=25

\(568-\left\{5.\left[143-\left(4-1\right)^2\right]+10\right\}\cdot1\)

\(\Rightarrow568-\left\{5\cdot\left[143-3^2\right]+10\right\}\)

\(\Rightarrow568-\left\{5\cdot\left\{143-9\right\}+10\right\}\)

\(\Rightarrow568-\left\{5\cdot134+10\right\}\)

\(\Rightarrow568-\left\{670+10\right\}\)

\(\Rightarrow568-680\)

\(\Rightarrow-112\)

a=4

b=0

( x - 2 )( 3 - x ) = 0

⇒ x - 2 = 0 hoặc 3 - x = 0

Nếu x - 2 = 0 ⇒ x = 0 + 2 = 2

Nếu 3 - x = 0 ⇒ x = 3 - 0 = 3

Vậy x ϵ { 2; 3 } để ( x - 2 )( 3 - x ) = 0

`(x-2)(3-x)=0`

`=> [(x-2=0),(3-x=0):}`

`=> [(x=2),(x=3):}`

Vậy `xin{2;3}`

Ta có 300 = 2² . 3 . 5²

90 = 2 . 3² . 5

144 = 2⁴ . 3²

Thừa số nguyên tố chung là 2 và 3

ƯCLN ( 300; 90; 144 ) = 2 . 3 = 6

Vậy ƯCLN ( 300; 90; 144 ) = 6