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

\(\Rightarrow3A=3^2+3^3+...+3^{50}+3^{51}\)

\(\Rightarrow3A-A=3^{51}-3\)

\(\Rightarrow2A=3^{51}-3\)

\(\Rightarrow A=\frac{3^{51}-3}{2}\)

\(B=2-2^2+2^3-2^4+...+2^{2019}-2^{2020}\)

\(2B=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)

\(B+2B=2-2^{2021}\)

\(3B=2-2^{2021}\)

\(B=\frac{2-2^{2021}}{3}\)

\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\)

\(C=1-\frac{1}{2009}\)

\(C=\frac{2008}{2009}\)

\(D=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(D=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(D=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(D=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)

= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)

= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)

=\(\dfrac{3}{2}.\dfrac{56}{305}\)

= \(\dfrac{78}{305}\)

\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)

*Nếu \(x^2-4=0\)

⇒ x² = 4

⇒ x ∈ {2 ; -2}

*Nếu \(6-2x=0\)

⇒2x = 6

⇒ x = 6 : 2 = 3

Vậy x ∈ { -2 ; 2 ; 3 }

a: =>5x=3x-6

=>2x=-6

hay x=-3

b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)

=>x-3=10 hoặc x-3=-10

=>x=13 hoặc x=-7

c: \(\left|x^3+1\right|+2\ge2\forall x\)

Dấu '=' xảy ra khi x=-1

\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)

\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)

\(=\frac{4}{3}.\frac{1}{3}+4\)

\(=4+4\)

\(=8\)

https://hoc247.net/hoi-dap/toan-6/chung-minh-s-1-2-2-2-2-3-2-4-2-5-2-6-2-7-chia-het-cho-3-faq250754.html

S= \(1+2+2^2+...+2^7\)

2S= \(2\cdot\left(2+2^2+...+2^7\right)\)

2S= \(2^1+2^2+...2^8\)

1S= 2S - S = \(\left(2^1+2^2+...2^8\right)-\left(1+2+2^2+...+2^7\right)\)

1S= \(2^1+2^2+...+2^8-1-2-2^2-...-2^7\)

1S= \(2^8-1\)

1S= \(256-1\)

1S= 255

=> 1S chia hết cho 3

Mà 1S= S

=> S chia hết cho 3

Vậy S chia hết cho 3

giải giúp mk với mk sắp đi học rồi

ai nhanh k 2 lan

 1)

a)-24+3(x-4)=111

           3(x-4)=111-(-24)

           3(x-4)=111+24

           3(x-4)=135

                x-4=135:3

                x-4=45

                 x  =45+4

                 x  =49

b)(2x-4)(3x+63)=0

\(\Rightarrow\)\(\orbr{\begin{cases}2x-4=0\\3x+63=0\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=2\\x=-21\end{cases}}\)

Vậy x\(\in\){2;-21}

c)|x-7|-4=(-2)⁴

   |x-7|    =(-2)⁴+4

   |x-7|    =16+4

   |x-7|    =20

\(\Rightarrow\)\(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=14\\x=0\end{cases}}\)

Vậy x\(\in\){14;0}

d)(x-1)²=144

    (x-1)²=12²

\(\Rightarrow\)x-1=12

          x  =12+1

          x  =13

e)(x+7)³=-8

   (x+7)³=(-2)³

\(\Rightarrow\)x+7=-2

         x     =-2-7

         x     =-9

2)

a)Ta có:

\(3n+12⋮n-3\)

\(\Rightarrow3n-9+21⋮n-3\)

\(\Rightarrow3\left(n-3\right)+21⋮n-3\)

\(\Rightarrow21⋮n-3\)

\(\Rightarrow n-3\inƯ\left(21\right)\)

\(\Rightarrow n-3\in\left\{1;3;7;21\right\}\)

Ta có bảng sau:

Vậy\(n\in\left\{4;6;10;24\right\}\)

b)Ta có:

\(n+9⋮n-1\)

\(\Rightarrow n-1+10⋮n-1\)

\(\Rightarrow10⋮n-1\)

\(\Rightarrow\)\(n-1\inƯ\left(10\right)\)

\(\Rightarrow n-1\in\left\{1;2;5;10\right\}\)

Ta có bảng sau:

Vậy \(n\in\left\{2;3;6;11\right\}\)