\(\dfrac{2^3+2^4+2^5}{7^2}=\dfrac{2^3\left(1+2+2^2\right)}{7^2}=\dfrac{8}{7}\)

\(\dfrac{2^3+2^4+2^5+2^6}{15^2}=\dfrac{2^3\left(1+2+2^2+2^3\right)}{15^2}=\dfrac{8}{15}\)

=\(\dfrac{3}{2}+\dfrac{1}{5}-\dfrac{3}{4}\)

=\(\dfrac{30}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)

=\(\dfrac{19}{20}\)

=\(\dfrac{4}{3}+\dfrac{2}{3}-\dfrac{3}{4}-\dfrac{4}{5}-\dfrac{6}{5}\)

=(\(\dfrac{4}{3}+\dfrac{2}{3}\)) -\(\dfrac{4}{3}-\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\)

=2\(-\dfrac{4}{3}-\dfrac{-2}{5}\)

=\(\dfrac{30}{15}-\dfrac{20}{15}-\dfrac{-6}{15}\)

=\(\dfrac{4}{15}\)

a) Ta có: ∠A + ∠B + ∠C = 180 độ

Mà ∠A=∠B=2∠C

⇒ 5∠C = 180 độ

⇒ ∠C = 180 độ : 5 = 36 độ

⇒ ∠A=∠B=2∠C=2.36 độ = 72 độ

Vậy ∠A=∠B=72 độ, ∠C=36 độ

b) Δ ABC là Δ nhọn

\(\dfrac{3}{8}:\left(\dfrac{7}{22}-\dfrac{2}{11}\right)+\dfrac{3}{8}:\left(\dfrac{2}{5}-\dfrac{1}{10}\right)\)

\(=\dfrac{3}{8}:\left(\dfrac{7}{22}-\dfrac{4}{22}\right)+\dfrac{3}{8}:\left(\dfrac{4}{10}-\dfrac{1}{10}\right)\)

\(=\dfrac{3}{8}:\left(\dfrac{7-4}{22}\right)+\dfrac{3}{8}:\left(\dfrac{4-1}{10}\right)\)

\(=\dfrac{3}{8}:\dfrac{3}{22}+\dfrac{3}{8}:\dfrac{3}{10}\)

\(=\dfrac{3}{8}.\dfrac{22}{3}+\dfrac{3}{8}.\dfrac{10}{3}\)

\(=\dfrac{3}{8}.\left(\dfrac{22}{3}+\dfrac{10}{3}\right)\)

\(=\dfrac{3}{8}.\left(\dfrac{22+10}{3}\right)\)

\(=\dfrac{3}{8}.\dfrac{32}{3}\)

\(=\dfrac{3.32}{8.3}\)

`=32/8`

`=4`