\(3A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{103}}-\dfrac{1}{3^{104}}\)

=>\(4A=3-\dfrac{1}{3^{104}}=\dfrac{3^{105}-1}{3^{104}}\)

=>\(A=\dfrac{3^{105}-1}{3^{104}\cdot4}\)

\(3A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{103}}-\dfrac{1}{3^{104}}\)

=>\(4A=3-\dfrac{1}{3^{104}}=\dfrac{3^{105}-1}{3^{104}}\)

=>\(A=\dfrac{3^{105}-1}{3^{104}\cdot4}\)

Bài 1:

a: \(\sqrt{20}-\sqrt{45}+3\sqrt{80}\)

\(=2\sqrt5-3\sqrt5+3\cdot4\sqrt5\)

\(=-\sqrt5+12\sqrt5=11\sqrt5\)

b: \(\frac{2}{3+\sqrt5}-\frac{2}{3-\sqrt5}\)

\(=\frac{2\left(3-\sqrt5\right)-2\left(3+\sqrt5\right)}{\left(3+\sqrt5\right)\left(3-\sqrt5\right)}\)

\(=\frac{6-2\sqrt5-6-2\sqrt5}{9-5}=\frac{-4\sqrt5}{4}=-\sqrt5\)

\(=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{50\cdot\dfrac{49}{2}}\)

\(=\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{50\cdot\dfrac{49}{2}}\)

\(=\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{49\cdot50}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

=2*24/50=48/50=24/25

ai giúp tui nhanh ik

Bài 2: 

a: \(\left(6x-39\right):7=3\)

\(\Leftrightarrow6x-39=21\)

hay x=10

Bạn kiểm tra lại đề nhé, hình như đề hơi có vấn đề

Ta có: A=3+3^2+3^3+....+3^2012

nên 3.A=3^2+3^3+3^4+....+3^2013

nên 3A-A=3^2013-3

nên 2A=3^2013-3

nên A=(3^2013-3)/2