\(a,=\dfrac{5}{7}-\dfrac{33}{8}=-\dfrac{191}{56}\\ b,=\left(\dfrac{12}{17}+\dfrac{5}{17}\right)+\left(\dfrac{19}{7}+\dfrac{3}{7}\right)=1+3=4\\ c,=\left(0,125\cdot8\right)^{12}-\left(\dfrac{45}{15}\right)^3=1-3^3=-26\\ d,=\left(-\dfrac{1}{3}\right)\left(5\dfrac{2}{7}-2\dfrac{2}{7}\right)=-\dfrac{1}{3}\cdot3=-1\\ e,=\dfrac{3^4\cdot3^6}{3^9}=3\)

a: \(=\dfrac{-2}{12}-\dfrac{5}{12}+\dfrac{7}{12}=0\)

b: \(=\left(\dfrac{4}{45}-\dfrac{1}{45}+\dfrac{7}{45}+\dfrac{4}{45}-\dfrac{2}{45}-\dfrac{9}{45}\right)=\dfrac{3}{45}=\dfrac{1}{15}\)

\(\dfrac{14}{7}>\dfrac{9}{7}\)

\(\dfrac{6}{13}< 1\)

\(\dfrac{7}{9}=\dfrac{14}{18}\)

\(\dfrac{4}{9}=\dfrac{20}{45}< \dfrac{26}{45}\)

\(\dfrac{11}{24}>\dfrac{9}{24}=\dfrac{3}{8}\)

\(\dfrac{42}{91}=\dfrac{42:7}{91:7}=\dfrac{6}{13}< \dfrac{7}{13}\)

\(\dfrac{34}{64}< \dfrac{52}{64}=\dfrac{13}{16}\)

\(\dfrac{17}{30}>\dfrac{16}{30}=\dfrac{8}{15}\)

\(\dfrac{9}{5}+\dfrac{5}{7}+\dfrac{7}{5}+\dfrac{3}{7}=\left(\dfrac{9}{5}+\dfrac{7}{5}\right)+\left(\dfrac{5}{7}+\dfrac{3}{7}\right)=\dfrac{16}{5}+\dfrac{8}{7}=\dfrac{112}{35}+\dfrac{40}{35}=\dfrac{152}{35}\)

\(\dfrac{1}{2}\times\dfrac{45}{33}\times\dfrac{1}{9}\times\dfrac{11}{6}=\dfrac{1}{2}\times\dfrac{15}{11}\times\dfrac{1}{9}\times\dfrac{11}{6}=\left(\dfrac{1}{2}\times\dfrac{1}{9}\right)\times\left(\dfrac{15}{11}\times\dfrac{11}{6}\right)=\dfrac{1}{18}\times\dfrac{15}{6}=\dfrac{5}{36}\)

Câu 1:

\(-\dfrac{5}{7}\cdot\dfrac{7}{-45}=\dfrac{1}{9}\) ; \(-\dfrac{3}{10}\cdot-30=9\) ; \(-12\cdot-\dfrac{2}{36}=\dfrac{2}{3}\)

Caau2;3;4;5,... tự bấm máy tính là ra

Ta thấy mẫu của dãy có dạng 1.5; 5.9; 9.13; 13.17; 17.21;... tổng quát là (4n-3)(4n+1). Mẫu thứ 100 bằng 397.401. Tổng của 100 số hạng đầu của dãy bằng:

\(\left(1-\dfrac{1}{401}\right):4=\dfrac{1}{4}-\dfrac{1}{1604}< \dfrac{1}{4}\)

a) \(\dfrac{4}{7}< \dfrac{6}{7}\)

\(\dfrac{15}{23}>\dfrac{12}{23}\)

\(\dfrac{45}{60}=\dfrac{3}{4}\)

b) \(\dfrac{8}{9}>\dfrac{11}{18}\)

\(\dfrac{13}{18}>\dfrac{2}{3}\)

\(\dfrac{27}{25}>1\)

hazz

a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)

\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)

\(=\dfrac{15}{26}-\dfrac{4}{26}\)

\(=\dfrac{11}{26}\)

b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)

\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)

\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)

\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)

\(=\dfrac{-967}{63}\)

c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)

\(=-1\)

d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)

=0