\(A=\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}=\dfrac{3}{7}\)

Ta có \(A=\frac{1}{3}.\frac{4}{6}.\frac{7}{9}.....\frac{208}{210}=\frac{1.4.7.....208}{3.6.9.....210}\)

Mà \(1< 3,4< 6,7< 9,...,208< 210\)

\(\Rightarrow1.4.7.....208< 3.6.9.....210\)

\(\Rightarrow\frac{1.4.7.....208}{3.6.9.....210}< 1\)\(\Leftrightarrow A< 1\)

Lại có \(1< 25\)\(\Rightarrow A< 25\)

\(\dfrac{1}{6}+\dfrac{-2}{5}=\dfrac{5}{30}+\dfrac{-12}{30}=\dfrac{-7}{30}\)

\(\dfrac{5}{12}-\dfrac{7}{8}=\dfrac{10}{24}-\dfrac{21}{24}=\dfrac{-11}{24}\)

\(\dfrac{-4}{25}:\dfrac{-32}{45}=\dfrac{-4}{25}.\dfrac{-45}{32}=\dfrac{-1}{5}.\dfrac{-9}{8}=\dfrac{4}{40}\)

\(\dfrac{-2}{9}.\dfrac{7}{11}+\dfrac{-2}{9}.\dfrac{10}{11}+\dfrac{-2}{9}.\dfrac{5}{11}=\dfrac{-2}{9}\left(\dfrac{7}{11}+\dfrac{10}{11}+\dfrac{5}{11}\right)=\dfrac{-2}{9}.\dfrac{22}{11}=\dfrac{-2}{9}.2=\dfrac{-4}{9}\)

a) \(\dfrac{11\cdot8-11\cdot3}{17-6}\)

\(=\dfrac{11\cdot\left(8-3\right)}{11}=5\)

b) \(\dfrac{24-12\cdot13}{12+4\cdot9}\)

\(=\dfrac{12\cdot\left(2-13\right)}{12\left(1+3\right)}=\dfrac{-11}{4}\)

a: \(=\dfrac{13\cdot\left(9-2\right)}{13}=7\)

b: \(=\dfrac{14\left(3-8\right)}{7\left(1+3\cdot3\right)}=2\cdot\dfrac{-5}{10}=-1\)

c: \(=\dfrac{54-72}{36}=\dfrac{-18}{36}=-\dfrac{1}{2}\)

d: \(=\dfrac{5^3}{10^2\cdot5}=\dfrac{5^2}{100}=\dfrac{1}{4}\)

a) (-12).10+9.10 = 10.[(-12)+9] = 10.(-3) = -30

b) (-11).8-(-11).9 = (-11).(8-9) = (-11).(-1) = 11

c) (-5).(-4) + (-5).14 = (-5).[(-4)+14] = -5.10 = -50

d) 13.(-3) + 13.8 = 13.[(-3)+8] = 13.5 = 65