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

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

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

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

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

k cho mik nha

A=120:{60:[(3mux2 + 4mux2)-5]}

A=120:{60[(9+16)-5]}

A=120:{60[25-5]}

A=120:{60.20}

A=120:1200

A=0,1

ko cos mays tính à

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

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

Sửa đề: Tính \(\left(S-P\right)^{2022}\) và \(P=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

Ta có: \(S=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=\left(1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2021}+\frac{1}{2022}\right)-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

=P

=>S-P=0

=>\(\left(S-P\right)^{2022}=0\)

2021^1=2021 đk ạ??

sigma

Ta có \(x+1=2022\)

\(P\left(x\right)=x^{101}-\left(x+1\right)x^{100}+...+\left(x+1\right)x-1\)

\(=x^{101}-x^{101}-x^{100}+...+x^2+x-1=x-1\)

-> P(x) = 2020