Bài 4 trang 30 toán lớp 7 tập 1 CDTính:a. $0,3 - \frac{4}{9}:\frac{4}{3}.\frac{6}{5}+1$b. $\left (...

a. $0,3 - \frac{4}{9}:\frac{4}{3}.\frac{6}{5}+1$

Cách 1:
$0,3 - \frac{4}{9}:\frac{4}{3}.\frac{6}{5}+1$
$=0,3 - \frac{4}{9} \times \frac{3}{4} \times \frac{6}{5} + 1$
$=0,3 - \frac{2}{5} + 1$
$=\frac{3}{10} - \frac{4}{10} + \frac{10}{10}$
$=\frac{9}{10}$

Cách 2:
$0,3 - \frac{4}{9}:\frac{4}{3}.\frac{6}{5}+1$
$=0,3 - \frac{4}{9} \times \frac{3}{4} \times \frac{6}{5} + 1$
$=0,3 - \frac{2}{5} + 1$
$=\frac{3}{10} - \frac{4}{10} + \frac{10}{10}$
$=\frac{9}{10}$

Câu trả lời: $\frac{9}{10}$

b. $\left ( \frac{-1}{3} \right )^{2} -\frac{3}{8}:(0,5)^{3}-\frac{5}{2}.(-4)$

Cách 1:
$\left ( \frac{-1}{3} \right )^{2} -\frac{3}{8}:(0,5)^{3}-\frac{5}{2}.(-4)$
$=\frac{1}{9} - \frac{3}{8} : \frac{1}{8} - \frac{5}{2} \times (-4)$
$=\frac{1}{9} - 3 + 10$
$=\frac{1}{9} - \frac{27}{9} + \frac{90}{9}$
$=\frac{64}{9}$

Cách 2:
$\left ( \frac{-1}{3} \right )^{2} -\frac{3}{8}:(0,5)^{3}-\frac{5}{2}.(-4)$
$=\frac{1}{9} - \frac{3}{8} : \frac{1}{8} - \frac{5}{2} \times (-4)$
$=\frac{1}{9} - 3 + 10$
$=\frac{1}{9} - \frac{27}{9} + \frac{90}{9}$
$=\frac{64}{9}$

Câu trả lời: $\frac{64}{9}$

c. $1 + 2:\left (\frac{2}{3}-\frac{1}{6}\right ).(-2,25)$

$= 1 + 2 : \left ( \frac{3}{4} - \frac{1}{6} \right ) \times (-2,25)$
$= 1 + 2 : \frac{9}{12} \times (-2,25)$
$= 1 + 2 \times \frac{12}{9} \times (-2,25)$
$= 1 + (-9) = -8$

Câu trả lời: $-8$

d. $\left [ \left ( \frac{1}{4} - 0,5 \right ).2 + \frac{8}{3} \right ]:2$

$= \left [ \left ( \frac{1}{4} - \frac{1}{2} \right ) \times 2 + \frac{8}{3} \right ] : 2$
$= \left ( \frac{-1}{4} \times 2 + \frac{8}{3} \right ) : 2$
$= \left ( \frac{-1}{2} + \frac{8}{3} \right ) : 2$
$= \frac{13}{6} \times \frac{1}{2} = \frac{13}{12}$

Câu trả lời: $\frac{13}{12}$