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`

M = (100 – 1).(100 – 22). (100 – 32)…(100 – 502)

M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92) .(100 – 102) .(100 – 112) …(100 – 502)

M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92). (100 – 100) .(100 – 112) …(100 – 502)

M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92) .0.(100 – 112) …(100 – 502)

M = 0

Vậy M = 0

100 = 10

100 - 10 ² = 0

M =( ...) ( ...) 0. ...( ...) = 0 

Lời giải:
\(\frac{3}{5}-[\frac{-1}{15}+\frac{16}{15}.(\frac{1}{4})^2]=\frac{3}{5}-[\frac{-1}{15}+\frac{16}{15}.\frac{1}{16}]\)

\(=\frac{3}{5}-(\frac{-1}{15}+\frac{1}{15})=\frac{3}{5}-0=\frac{3}{5}\)

62/12

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=\dfrac{x+2y+z}{5+2.4+7}=\dfrac{10}{20}=\dfrac{1}{2}\\ =>x=\dfrac{5}{2};y=\dfrac{4}{2}=2;z=\dfrac{7}{2}\)