Giải SBT Toán 7 Bài 4. Thứ tự thực hiện phép tính. Quy tắc dấu ngoặc có đáp án

Lời giải:

$12\,.\,{\left( {\frac{2}{3} - \frac{5}{6}} \right)^2} = 12\,.\,{\left( {\frac{4}{6} - \frac{5}{6}} \right)^2}$

$= 12\,.\,{\left( {\frac{{ - 1}}{6}} \right)^2} = 12\,.\,\frac{1}{{36}} = \frac{1}{3}$.

Lời giải:

$4\,.\,{\left( { - \frac{1}{2}} \right)^3} - 2.{\left( { - \frac{1}{2}} \right)^2} + 3\,.\,\left( { - \frac{1}{2}} \right) + 1$.

$= 4\,.\,\frac{{ - 1}}{8} - 2.\frac{1}{4} + \frac{{ - 3}}{2} + 1$

$= \frac{{ - 1}}{2} - \frac{1}{2} + \frac{{ - 3}}{2} + 1$

$= - 1 + 1 + \frac{{ - 3}}{2}$

$= 0 + \frac{{ - 3}}{2} = \frac{{ - 3}}{2}$.

Lời giải:

$$\frac{{45}}{4} - \left( {2\frac{5}{7} + 5{,}25} \right) = \frac{{45}}{4} - \left( {\frac{{19}}{7} + \frac{{21}}{4}} \right)$$

$$= \frac{{45}}{4} - \frac{{19}}{7} - \frac{{21}}{4} = \frac{{45}}{4} - \frac{{21}}{4} - \frac{{19}}{7}$$

$= 6 - \frac{{19}}{7} = \frac{{42}}{7} - \frac{{19}}{7} = \frac{{23}}{7}$;

Lời giải:

$\frac{5}{9}:2{,}4 - \frac{{41}}{9}:2{,}4 = \frac{5}{9}:\frac{{12}}{5} - \frac{{41}}{9}:\frac{{12}}{5}$

$= \frac{5}{9}.\frac{5}{{12}} - \frac{{41}}{9}.\frac{5}{{12}} = \left( {\frac{5}{9} - \frac{{41}}{9}} \right).\frac{5}{{12}}$

$= - 4.\frac{5}{{12}} = - \frac{5}{3}$;

Lời giải:

$\left( { - \frac{3}{4} + \frac{5}{{13}}} \right).\,\frac{7}{2} - \left( {\frac{9}{4} - \frac{8}{{13}}} \right).\,\frac{7}{2}$

$= \left[ { - \frac{3}{4} + \frac{5}{{13}} - \left( {\frac{9}{4} - \frac{8}{{13}}} \right)} \right].\,\frac{7}{2}$

$= \left[ { - \frac{3}{4} + \frac{5}{{13}} - \frac{9}{4} + \frac{8}{{13}}} \right].\,\frac{7}{2}$

$= \left[ {\left( { - \frac{3}{4} - \frac{9}{4}} \right) + \left( {\frac{5}{{13}} + \frac{8}{{13}}} \right)} \right].\,\frac{7}{2}$

$= \left( { - 3 + 1} \right).\,\frac{7}{2} = ( - 2).\,\frac{7}{2} = - 7$.