Giải bài 3 trang 25 toán 7 tập 1 cánh diều

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{36}{169}$

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