C : 4

Ta có: \(\dfrac{2}{5}=\dfrac{a}{10}\) \(\Leftrightarrow\dfrac{4}{10}=\dfrac{a}{10}\) \(\Rightarrow a=4\)

-> C: 4

C. x = 4

a) \(\dfrac{3}{4}=\dfrac{15}{20}\)

b) \(\dfrac{4}{5}=\dfrac{12}{15}\)

c) \(\dfrac{-4}{9}=\dfrac{-16}{36}\)

d) \(\dfrac{7}{-13}=\dfrac{21}{-39}\)

a ) \(\dfrac{3}{4}\)= \(\dfrac{15}{20}\)

b )\(\dfrac{4}{5}\)= \(\dfrac{12}{15}\)

c) \(\dfrac{-4}{9}\)=\(\dfrac{-16}{36}\)

d) \(\dfrac{7}{-13}\)=\(\dfrac{21}{-39}\)

\(\dfrac{7}{21}+\dfrac{-9}{36}=\dfrac{1}{3}+\dfrac{-1}{4}=\dfrac{4}{12}+\dfrac{-3}{12}=\dfrac{1}{12}\)

\(\dfrac{-12}{18}+\dfrac{-21}{35}=\dfrac{-2}{3}+\dfrac{-3}{5}=\dfrac{-10}{15}+\dfrac{-9}{15}=\dfrac{-19}{15}\)

\(\dfrac{-18}{14}+\dfrac{15}{-21}=\dfrac{-9}{7}+\dfrac{-5}{7}=\dfrac{-14}{7}=-2\)

\(\dfrac{3}{21}+\dfrac{-6}{42}=\dfrac{1}{7}+\dfrac{-1}{7}=0\)

\(a)\dfrac{5}{7}\times\dfrac{7}{15}\\ =\dfrac{5\times7}{7\times15}\\ =\dfrac{35}{105}=\dfrac{1}{3}\\ b)\dfrac{8}{21}\times3\\ =\dfrac{8\times3}{21}\\ =\dfrac{24}{21}=\dfrac{8}{7}\\ c)\dfrac{24}{17}:\dfrac{9}{17}\\ =\dfrac{24}{17}\times\dfrac{17}{9}\\ =\dfrac{24\times17}{17\times9}\\ =\dfrac{408}{153}=\dfrac{8}{3}\\ d)\dfrac{36}{5}:9\\ =\dfrac{36}{5}\times\dfrac{1}{9}\\ =\dfrac{36\times1}{5\times9}\\ =\dfrac{36}{45}=\dfrac{4}{5}\)

a) \(\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5\times7}{7\times15}=\dfrac{35}{105}=\dfrac{1}{3}\)

b) \(\dfrac{8}{21}\times3=\dfrac{8\times3}{21}=\dfrac{24}{21}=\dfrac{8}{7}\)

c) \(\dfrac{24}{17}:\dfrac{9}{17}=\dfrac{24}{17}\times\dfrac{17}{9}=\dfrac{24\times17}{17\times9}=\dfrac{24}{9}=\dfrac{8}{3}\)

d) \(\dfrac{36}{5}:9=\dfrac{36}{5}\times\dfrac{1}{9}=\dfrac{36}{5\times9}=\dfrac{36}{45}=\dfrac{4}{5}\)

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

Cần đáp án thoi

A

a) \(\dfrac{-12}{15}+\dfrac{-4}{26}=\dfrac{-4}{5}+\dfrac{-2}{13}=\dfrac{-52-10}{65}=\dfrac{-62}{65}\)

b) \(5\dfrac{1}{3}-2\dfrac{4}{5}=\dfrac{16}{3}-\dfrac{14}{5}=\dfrac{80}{15}-\dfrac{42}{15}=\dfrac{38}{15}\)

c) \(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)+\dfrac{-5}{10}=\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{1}{2}=\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{35}{70}=\dfrac{41}{70}\)

d) \(-1\dfrac{2}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-9}{7}+\dfrac{3}{14}-\dfrac{5}{21}=\dfrac{-54}{42}+\dfrac{9}{42}-\dfrac{10}{42}=\dfrac{-55}{42}\)

e) \(12-\dfrac{11}{121}+\left(\dfrac{-8}{9}\right)-\left(-\dfrac{3}{7}\right)\)

\(=12-\dfrac{11}{121}-\dfrac{8}{9}+\dfrac{3}{7}\)

\(=\dfrac{91476}{7623}-\dfrac{693}{7623}-\dfrac{6776}{7623}+\dfrac{3267}{7623}\)

\(=\dfrac{7934}{693}\)

a: \(\Leftrightarrow-4< =x< =-3\)

hay \(x\in\varnothing\)

b: =>-9<x<=3

hay \(x\in\left\{0;1;2;3\right\}\)

a: \(\dfrac{5}{6}\cdot7=\dfrac{5\cdot7}{6}=\dfrac{35}{6}\)

b: \(\dfrac{7}{10}\cdot3=\dfrac{7\cdot3}{10}=\dfrac{21}{10}\)

c: \(5\cdot\dfrac{4}{21}=\dfrac{5\cdot4}{21}=\dfrac{20}{21}\)

d: \(2\cdot\dfrac{5}{9}=\dfrac{2\cdot5}{9}=\dfrac{10}{9}\)