\(2n+9⋮3n+1\)

\(\Rightarrow3\left(2n+9\right)⋮3n+1\)

\(\Rightarrow2\left(3n+1\right)+25⋮3n+1\)

\(\Rightarrow25⋮3n+1\)

\(\Rightarrow3n+1\in\left\{5,25,1,-5,-25,-1\right\}\)

\(n\in\left\{8,0\right\}\)

\(5n+2⋮9-2n\)

\(\Rightarrow2\left(5n+2\right)⋮9-2n\)

\(\Rightarrow-5\left(9-2n\right)-41⋮9-2n\)

\(41⋮9-2n\)

\(\Rightarrow9-2n\in\left\{41,-41,1,-1\right\}\)

\(\Rightarrow n\in\left\{-16,25,4,-5\right\}\)

34+35+36+37-24-25-26-27

= (34-24)+(35-25)+(36-26)+(37-27)

= 10+10+10+10

= 40

tính tổng sau một cách hợp lý

34+35+36+37-24-25-26-27

=(34-24)+(35-25)+(36-26)+(37-27)

=10+10+10+10

=40

Câu a = 2

a) \(2x+\left(5x+11\right)=25\)

\(2x+5x+11=25\)

\(7x=14\)

\(x=2\)

b) \(3^{2x}\cdot3=3^{2020}:3^{2017}\)

\(3^{2x}\cdot3=3^3\)

\(3^{2x}=3^2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

Vậy........

\(10-2n⋮n-2\)

\(-2\left(n-2\right)+6⋮n-2\)

\(6⋮n-2\)

\(n-2\in\left\{6,2,1,3,-6,-2,-1,-3\right\}\)

\(n\in\left\{8,4,3,5,-4,0,1,-1\right\}\)

\(306-216:18\)

\(=306-12\)

\(=294\)

\(13\cdot34+66\cdot\left|-13\right|-1338\)

\(=13\left(34+66\right)-1338\)

\(=-38\)

 1-2-3+4+5-6-...+2009-2010

=(1-2-3+4)+(5-6-7+8)+...+(2009-2010)

=    0        +      0       +....+    (-1)

=      -1

\(1-2-3+4+5-6-.....+2009-2010\)

\(=1-2-3+4+5-6-7+8+.....+2006-2007-2008+2009-2010\)(bước này là mk viết rõ ra thôi)

\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2006-2007-2008+2009\right)-2010\)

\(=0+0+......+0-2010\)

\(=-2010\)

~ học tốt ~

nhớ  

Ở chỗ (2x=1 ) nghĩa là j zậy bạn ????

\(24-x-1=17-\left(2x-1\right)\)

\(23-13-1=-2x-x\)

\(9=-x\)

\(x=-9\)

A=4+4^2+4^3+4^4+...+4^49+4^50

A=(4+4^2)+(4^3+4^4)+...+(4^49+4^50)

A=4.(1+4)+4^3.(1+4)+...+4^49.(1+4)

A=4.5+4^3.5+...+4^49.5

A=5.(4+4^3+...+4^49) chia het cho 5(vi 5 chia het cho 5)

=> A chia het cho 5

\(A=4+4^2+4^3+4^4+...+4^{49}+4^{50}\)

\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{49}+4^{50}\right)\)

\(A=4.5+4^3.5+...+4^{49}.5\)

\(A=5.\left(4+4^3+...+4^{49}\right)CHIA-HETCHO5\)