CHẳng hỉu j

Ngũ hay mũ bạn ???

Ai giúp mình với , câu hỏi nâng cao

TA có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)

\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)

TA có: \(20^{10}-1>20^{10}-3\)

=>\(\frac{2}{20^{10}-1}<\frac{2}{20^{10}-3}\)

=>\(\frac{2}{20^{10}-1}+1<\frac{2}{20^{10}-3}+1\)

=>A<B

\(A=\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\)

\(2A=1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\)

\(2A-A=\)\(\left(1-\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3+...-\left(\frac{1}{2}\right)^{19}\right)-\)\(\left(\frac{1}{2}-\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3-\left(\frac{1}{2}\right)^4+...-\left(\frac{1}{2}\right)^{20}\right)\)

\(A=1-\left(\frac{1}{2}\right)^{20}\)

\(a=4^5.9^4-2.\dfrac{6^9}{2^{10}}.3^8+6^8.20\)

Đề là như vầy đúng ko bn?

$\textbf{A)}$

$\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}$

$=\dfrac{(2^2)^5(3^2)^4-2(2\cdot3)^9}{2^{10}\cdot3^8+2^2\cdot5(2\cdot3)^8}$

$=\dfrac{2^{10}3^8-2^{10}3^9}{2^{10}3^8+2^{10}\cdot5\cdot3^8}$

$=\dfrac{2^{10}3^8(1-3)}{2^{10}3^8(1+5)}$

$=\dfrac{-2}{6}$

$=-\dfrac{1}{3}.$

gấp lắm các bạn ơi đấy là bài cuối

M =2020