(45-x):6 = 18

45-x = 18 . 6

45-x = 108

x = 45 - 108

x = -63

45-3x = 35

3x = 45 - 15

3x = 30

x = 30 : 3

x = 10

5225:9.[215-(3x-18)] = 25

9.[215-(3x-18)] = 5225 : 25

9.[215-(3x-18)] = 209

215-(3x-18) = 209 : 9

215-(3x-18) = 209/9

3x-18 = 215 - 209/9

3x-18 = 1726/9

3x = 1726/9 + 18

3x = 1888/9

x = 1888/9 : 3

x = 1888/27

Ta có : x:5 = 30

x = 30 . 5

x = 150

Ta có :  gọi số bị chia là a , thương là b

410 : a = b ( dư 19 )

=> 391 chia hết cho a 

Mình chưa nghĩa ra :b

a+6=4x9

a+6=36  a= 36-6=30

bx5=100-75

bx5=25 b=25:5=5

Thương của phép chia a chia cho b là :

   30:5= 6

         Đáp số: 6

= 1,2 bạn nhé

\(=-\frac{\left(x^2+y^2\right)^4}{\left(x^2+y^2\right)^2}-\frac{4\left(x^2+y^2\right)^3}{\left(x^2+y^2\right)^2}-\frac{5\left(x^2+y^2\right)^2}{\left(x^2+y^2\right)^2}=-\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)-5\)

\(=-1-\left(\left(x^2+y^2\right)^2+4\left(x^2+y^2\right)+4\right)=-1-\left(x^2+y^2+2\right)^2\le-1< 0\forall x\left(đpcm\right).\)

B1

B = 5² . 4 - ( 18 + 6 . 7 ) : 81 : 3³

= 25 . 4 - ( 18 + 42 ) : 3⁴ : 3³

= 100 - 60 : 3

= 100 - 20

= 80

B2

5^x+1 + 52 = 6² + ( 7⁹ : 7⁷ - 2³ )

=> 5^x+1 + 52 = 36 + ( 7² - 8 )

=> 5^x+1 + 52 = 36 + 41

=> 5^x+1 + 52 = 77

=> 5^x+1 = 25

=> 5^x+1 = 5²

=> x + 1 = 2

=> x = 1

\(+)18⋮x-3\)

\(\Rightarrow x-3\inƯ\left(18\right)\)

mà \(Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)

\(\Rightarrow\hept{\begin{cases}x-3=1;x-3=6\\x-3=2;x-3=9\\x-3=3;x-3=18\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=4;x=9\\x=5;x=12\\x=6;x=21\end{cases}}\)

\(26⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\inƯ\left(26\right)\)

mà \(Ư\left(26\right)=\left\{1;2;13;26\right\}\)

\(\Rightarrow\orbr{\begin{cases}x+1=1\\x+1=2\end{cases}}\orbr{\begin{cases}x+1=13\\x+1=26\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\orbr{\begin{cases}x=12\\x=25\end{cases}}\)

Bài 1:

a: =>x^3-x-6x-6=0

=>x(x-1)(x+1)-6(x+1)=0

=>(x+1)(x-3)(x+2)=0

hay \(x\in\left\{-1;3;-2\right\}\)

b: \(\Leftrightarrow x^2-6x+9+y^2+6y+9=0\)

=>(x-3)^2+(y+3)^2=0

=>x=3 và y=-3

\(x^3-7x-6=0\)

\(x^3-3x^2+3x^2+2x-9x-6=0\)

\(x^2.\left(x-3\right)+3x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\left(x+3\right).\left(x^2+3x+2\right)=0\Rightarrow\left(x-3\right).\left(x^2+3x+x+2\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x+1\right).\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\text{hoặc }x=-2\)