Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)

\(=3\left(1-3+3^2-3^3+\cdots+3^{22}-3^{23}\right)\) ⋮3

Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)

\(=3\left(1-3\right)+3^3\left(1-3\right)+\cdots+3^{23}\left(1-3\right)\)

\(=3\cdot\left(-2\right)+3^3\cdot\left(-2\right)+\cdots+3^{23}\left(-2\right)\)

\(=-2\left(3+3^3+\cdots+3^{23}\right)\)

\(=-2\left\lbrack\left(3+3^3\right)+\left(3^5+3^7\right)+\cdots+\left(3^{21}+3^{23}\right)\right\rbrack\)

\(=-2\cdot\left\lbrack3\cdot\left(1+3^2\right)+3^5\left(1+3^2\right)+\cdots+3^{21}\left(1+3^2\right)\right\rbrack\)

\(=-2\cdot10\cdot\left(3+3^5+\cdots+3^{21}\right)=-20\cdot\left(3+3^5+\cdots+3^{21}\right)\) ⋮20

Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)

\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+\cdots-\left(3^{22}-3^{23}+3^{24}\right)\)

\(=3\left(1-3+3^2\right)-3^4\left(1-3+3^2\right)+\cdots+3^{22}\left(1-3+3^2\right)\)

\(=7\cdot\left(3-3^4+\cdots-3^{22}\right)\) ⋮7

Ta có: C⋮20

C⋮7

C⋮3

mà ƯCLN(20;3;7)=1

nên C⋮20*3*7

=>C⋮420

\(A=\)\(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)

\(3A=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{49}}-\frac{1}{3^{50}}\)

\(4A=-1-\frac{1}{3^{51}}\)

\(A=\frac{-1-\frac{1}{3^{51}}}{4}\)

k cho mik nha

9²:3³=(3²)²:3³=3⁴:3³=3^4–3=3

5².25²= 5².(5²)²=5².5⁴=5²⁺⁴=5⁶

a)9² : 3³ = (3²)² : 3³ = 3⁴ : 3³ = 3.

b) 5² . 25² = 5² . (5²)² = 5² . 5⁴ = 5^6.

c) \(\left(\frac{1}{3}\right)^2\) . \(\left(\frac{1}{9.3}\right)^2\) = \(\frac{1^2}{3^2}\). \(\frac{1^2}{27^2}\)= \(\frac{1}{9}\).\(\frac{1}{729}\)= \(\frac{1}{2511}\)

\(C=3-3^2+3^3-3^4+3^5-3^6+...-3^{22}+3^{23}-3^{24}\)

\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+...-\left(3^{22}-3^{23}+3^{24}\right)\)

\(=3\left(1-3+3^2\right)-3^4\left(1-3+3^2\right)+...-3^{22}\left(1-3+3^2\right)\)

\(=7\left(3-3^4+...-3^{22}\right)⋮7\)

\(C=3-3^2+3^3-3^4+3^5-3^6+...-3^{22}+3^{23}-3^{24}\)

\(=\left(3-3^2+3^3-3^4\right)+\left(3^5-3^6+3^7-3^8\right)+...+\left(3^{21}-3^{22}+3^{23}-3^{24}\right)\)

\(=3\left(1-3+3^2-3^3\right)+3^5\left(1-3+3^2-3^3\right)+...+3^{21}\left(1-3+3^2-3^3\right)\)

\(=-20\cdot\left(3+3^5+...+3^{21}\right)\)

\(=-60\cdot\left(1+3^4+...+3^{20}\right)⋮60\)

\(C⋮60;C⋮7\)

mà ƯCLN(60;7)=1

nên C chia hết cho 60*7=420

Nỏ biết hỏi lắm hỏi cấy lò tôn

Lời giải:

a. Biểu thức $B$ không có GTLN bạn nhé. Chỉ có GTNN thôi.

b. 

$C=(3-3^2+3^3-3^4)+(3^5-3^6+3^7-3^8)+....+(3^{21}-3^{22}+3^{23}-3^{24})$

$=(3-3^2+3^3-3^4)+3^4(3-3^2+3^3-3^4)+....+3^{20}(3-3^2+3^3-3^4)$

$=(3-3^2+3^3-3^4)(1+3^4+...+3^{20})=-60(1+3^4+...+3^{20})\vdots 60(*)$

Mặt khác:

$C=(3-3^2+3^3)-(3^4-3^5+3^6)+.....-(3^{22}-3^{23}+3^{24})$

$=3(1-3+3^2)-3^4(1-3+3^2)+...-3^{22}(1-3+3^2)$

$=(1-3+3^2)(3-3^4+...-3^{22})=7(3-3^4+...-3^{22})\vdots 7(**)$

Từ $(*); (**)$ mà $(7,60)=1$ nên $C\vdots (7.60)$ hay $C\vdots 420$

cảm ơn bạn nhé^^

  C = 3 - 3² + 3³ - 3⁴ + 3⁵ - 3⁶ +...+ 3²³ - 3²⁴

3C =      3² - 3³ + 3⁴ - 3⁵ + 3⁶-...- 3²³ + 3²⁴ - 3²⁵

3C - C = -3²⁵ - 3

2C      = -3²⁵ - 3

2C = - ( 3²⁵ + 3) = - [(3⁴)⁶. 3 + 3] = - [\(\overline{...1}\)⁶.3+3] = -[ \(\overline{..3}\)  + 3]

2C = - \(\overline{..6}\)

⇒ \(\left[{}\begin{matrix}C=\overline{..3}\\C=\overline{..8}\end{matrix}\right.\) 

⇒ C không thể chia hết cho 420 ( xem lại đề bài em nhé)

b, (\(x+1\))²⁰²² + (\(\sqrt{y-1}\) )²⁰²³ = 0

Vì (\(x+1\))²⁰²² ≥ 0 

\(\sqrt{y-1}\) ≥ 0 ⇒ (\(\sqrt{y-1}\))²⁰²³ ≥ 0

Vậy (\(x\) + 1)²⁰²² + (\(\sqrt{y-1}\))²⁰²³ = 0

⇔ \(\left\{{}\begin{matrix}\left(x+1\right)^{2022}=0\\\sqrt{y-1}=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x+1=0\\y-1=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

Kết luận: cặp (\(x,y\)) thỏa mãn đề bài là:

(\(x,y\)) = (-1; 1)