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\(A=3+3^2+...+3^{2005}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2006}\)

\(\Rightarrow3A-A=3^{2006}-3\)

\(\Rightarrow2A=3^{2006}-3\)

\(\Rightarrow2A+3=3^{2006}\) là 1 lũy thừa của 3 (đpcm)

4.

\(B=1+1+2+2^2+2^3+...+2^{100}\)

\(2B=2+2+2^2+...+2^{101}\)

\(\Rightarrow2B-B=2+2^{101}-\left(1+1\right)=2^{101}\)

\(\Rightarrow B=2^{101}\) là 1 lũy thừa của 2 (đpcm)

Bài 1:

\(A=2+2^2+2^3+...+2^{2003}+2^{2004}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2002}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+2^4+...+2^{2002}\right)⋮7\)

Bài 2:

\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\cdot\left(2+2^5+...+2^{57}\right)⋮15\)

Bài 3:

\(A=1+3+3^2+3^3+...+3^{1990}+3^{1991}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{1989}+3^{1990}+3^{1991}\right)\)

\(=13+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)

Bài 4:

\(A=4+4^2+4^3+4^4+...+4^{23}+4^{24}\)

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

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

\(=20\left(1+4^2+...+4^{22}\right)⋮20\)

Bài 33:

a: \(x^2-3x+2=\left(x-2\right)\left(x-1\right)\)

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

=\(-\dfrac{5}{3}\)

~ - 5/3 nhaa

đề bài: rút gọn biểu thức a

\(A=\left(x-3\right)^2-\left(2x+1\right)^2\)

\(=\left(x-3-2x-2\right)\left(x-3+2x+1\right)\)

\(=-\left(x+5\right)\left(3x-2\right)\)

À đề là rút gọn biểu thức thì:

\(A=\left(x-3\right)^2-\left(2x+1\right)^2\)

\(=x^2-6x+9-4x^2-4x-1=-3x^2-10x+8\)

9 . field