Đáp án:
a) `(``\frac{-2}{5}` + `\frac{1}{4}` `)` : `(``1` -`\frac{2}{5}``)`
= `\frac{-3}{20}` : `\frac{3}{5}`
= `\frac{-3}{20}` . `\frac{5}{3}`
= `\frac{-1}{4}`
b) `\frac{3}{4}` - `1``\frac{1}{2}` + `0,5` : `\frac{5}{12}`
= `\frac{3}{4}` - `\frac{3}{2}` + `\frac{1}{2}` : `\frac{5}{12}`
= `\frac{3}{4}` - `\frac{3}{2}` + `\frac{6}{5}`
= `\frac{15}{20}` - `\frac{30}{20}` + `\frac{24}{20}`
= `\frac{9}{20}`
c) `\frac{1}{5}` . `\frac{4}{7}` + `\frac{1}{5}` : `\frac{7}{10}` - `\frac{1}{5}`
= `\frac{1}{5}` . `\frac{4}{7}` + `\frac{1}{5}` . `\frac{10}{7}` - `\frac{1}{5}`
= `\frac{1}{5}` . `\frac{4}{7}` + `\frac{1}{5}` . `\frac{10}{7}` - `\frac{1}{5}` . `1`
= `\frac{1}{5}` . `(``\frac{4}{7}` + `\frac{10}{7}` - `1``)`
= `\frac{1}{5}` . `1`
= `\frac{1}{5}`
c) $(-2)^{2}$ - `1``\frac{5}{27}` . `(``\frac{-3}{2}^{3}``)`
= `4` - `\frac{32}{27}` . `(``\frac{-27}{8}``)`
= `4` - `(``-4``)`
= `4` + `4`
= `8`
