Giải bài 3 trang 25 toán 7 tập 1 cánh diều

a. $(3, 9 + \frac{1}{2} .2, 6) : 13$

$= (\frac{39}{10} + \frac{1}{2} . \frac{13}{5}) : 13$

$= (\frac{39}{10} + \frac{26}{10}) . \frac{1}{13}$

$= \frac{65}{10} . \frac{1}{13} = \frac{1}{2}$

$3, 9 + \frac{1}{2} .2, 6 : 13$

$= \frac{39}{10} + \frac{1}{2} . \frac{13}{5} . \frac{1}{13}$

$= \frac{39}{10} + \frac{1}{10} = \frac{40}{10} = 4$

=> $(3, 9 + \frac{1}{2} .2, 6) : 13 \neq 3, 9 + \frac{1}{2} .2, 6 : 13$

b. $\frac{5}{4} - {(\frac{1}{4})}^{2} = \frac{20}{16} - \frac{1}{16} = \frac{19}{16}$

${(\frac{5}{4} - \frac{1}{4})}^{2} = 1^{2} = 1$

=> $\frac{5}{4} - {(\frac{1}{4})}^{2} \neq {(\frac{5}{4} - \frac{1}{4})}^{2}$

c. $\frac{28}{9} .0, 7 + \frac{28}{9} .0, 5 = \frac{28}{9} . (0, 7 + 0, 5)$

=> $\frac{28}{9} .0, 7 + \frac{28}{9} .0, 5 = \frac{28}{9} . (0, 7 + 0, 5)$

d. $\frac{36}{13} : 4 + \frac{36}{13} : 9$

$= \frac{36}{13} . \frac{1}{4} + \frac{36}{13} . \frac{1}{9}$

$= \frac{36}{13} . (\frac{1}{4} + \frac{1}{9})$

$= \frac{36}{13} . \frac{13}{36} = 1$

$\frac{36}{13} : (4 + 9)$

$= \frac{36}{13} : 13$

$= \frac{36}{13} . \frac{1}{13}$

$= \frac{36}{169}$

=> $\frac{36}{13} : 4 + \frac{36}{13} : 9 \neq \frac{36}{13} : (4 + 9)$