a) Ta có: \(7\cdot\dfrac{3}{14}-\dfrac{1}{14}\)

\(=\dfrac{3}{2}-\dfrac{1}{14}\)

\(=\dfrac{21}{14}-\dfrac{1}{14}\)

\(=\dfrac{10}{7}\)

b) Ta có: \(\dfrac{3}{2}+\dfrac{7}{4}:\dfrac{5}{2}\)

\(=\dfrac{3}{2}+\dfrac{7}{4}\cdot\dfrac{2}{5}\)

\(=\dfrac{3}{2}+\dfrac{7}{10}\)

\(=\dfrac{15}{10}+\dfrac{7}{10}=\dfrac{22}{10}=\dfrac{11}{5}\)

Lời giải:
\(7\times \frac{3}{14}-\frac{1}{14}=\frac{7\times 3}{14}-\frac{1}{14}=\frac{21}{14}-\frac{1}{14}=\frac{21-1}{14}=\frac{20}{14}=\frac{2\times 10}{2\times 7}=\frac{10}{7}\)

\(\frac{3}{2}+\frac{7}{4}:\frac{5}{2}=\frac{3}{2}+\frac{7}{4}\times \frac{2}{5}=\frac{3}{2}+\frac{7\times 2}{4\times 5}=\frac{3}{2}+\frac{7\times 2}{2\times 2\times 5}\)

\(=\frac{3}{2}+\frac{7}{2\times 5}=\frac{3\times 5}{2\times 5}+\frac{7}{2\times 5}=\frac{3\times 5+7}{2\times 5}=\frac{22}{2\times 5}=\frac{2\times 11}{2\times 5}=\frac{11}{5}\)

a. 7 x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{7}{1}\) x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{98}{14}\) x \(\dfrac{3}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{3}{2}\) - \(\dfrac{1}{14}\)

= \(\dfrac{21}{14}\) - \(\dfrac{1}{14}\)

= \(\dfrac{10}{7}\)

b. \(\dfrac{3}{2}\) + \(\dfrac{7}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{6}{4}\) + \(\dfrac{7}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{13}{4}\) : \(\dfrac{5}{2}\)

= \(\dfrac{13}{4}\) . \(\dfrac{2}{5}\)

= \(\dfrac{13}{10}\)

=>x-1/2=3+1/14-5-1/3-4/7+11/21

=>x-1/2=-97/42

=>x-21/42=-97/42

=>x=-97/42+21/42=-76/42=-38/21

a)=2

b)=1/3

a,2

b,1/3

\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14

\(=\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{98-36}{168}=\dfrac{62}{168}=\dfrac{31}{84}\)

\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)

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

Ta có: \(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}.\dfrac{858585}{313131}.\left(-1\dfrac{14}{17}\right)\)

             \(=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4\left(1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}\right)}.\dfrac{85.10101}{32.10101}.\dfrac{-31}{17}=\dfrac{1}{4}.\dfrac{85}{31}.\dfrac{-31}{17}=-\dfrac{5}{4}\)

\(4\frac{2}{7}.3=\left(4.3\right)+\left(\frac{2}{7}.3\right)=12+\frac{6}{7}=12\frac{6}{7}\)

tk nha

a ( ....< ....)

b( ....<....)

a <

b <

tick cho mk nha

a) Áp dụng t/c dtsbn:

 \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

b) \(\dfrac{3}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)

Và \(x+16=y\Rightarrow y-x=16\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{y-x}{7-3}=\dfrac{16}{4}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.3=12\\y=4.7=28\end{matrix}\right.\)

a) \(\dfrac{x}{7}=\dfrac{y}{13}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

⇒\(\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)