`@` `\text {Ans}`
`\downarrow`
`1)`
\(A=\dfrac{5}{7}+\dfrac{3}{4}+\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{1}{4}\)
`=`\(\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{1}{5}\)
`=`\(1+1+\dfrac{1}{5}=\dfrac{11}{5}\)
Vậy, `A=`\(\dfrac{11}{5}\)
`2)`
\(B=4\dfrac{4}{9}\div\dfrac{8}{3}+\dfrac{2}{3}\div2\dfrac{2}{3}\)
`=`\(\dfrac{40}{9}\times\dfrac{3}{8}+\dfrac{2}{3}\times\dfrac{3}{8}\)
`=`\(\dfrac{3}{8}\times\left(\dfrac{40}{9}+\dfrac{2}{3}\right)\)
`=`\(\dfrac{3}{8}\times\dfrac{46}{9}=\dfrac{23}{12}\)
Vậy, `B =`\(\dfrac{23}{12}\)
`3)`
\(C=\dfrac{2}{15}\times\dfrac{13}{27}+\dfrac{2}{15}\times\dfrac{14}{27}\)
`=`\(\dfrac{2}{15}\times\left(\dfrac{13}{27}+\dfrac{14}{27}\right)\)
`=`\(\dfrac{2}{15}\times1=\dfrac{2}{15}\)
Vậy, `C=`\(\dfrac{2}{15}\)
`4)`
\(D=40,35+112,37-18,35+17,63\)
`=`\(\left(40,35-18,35\right)+\left(112,37+17,63\right)\)
`= 22+130`
`= 152`
Vậy, `D = 152.`
