Giải câu 96 bài: Ôn tập chương 1 sgk Toán 7 tập 1 Trang 48
Câu 96: trang 48 sgk toán lớp 7 tập 1
Thực hiện phép tính (bằng cách hợp lí nếu có thể):
a. $1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}$;
b. $\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}$;
c. $9.\left ( -\frac{1}{3} \right )^{3}+\frac{1}{3}$;
d. $15\frac{1}{4}\div \left ( -\frac{5}{7} \right )-25\frac{1}{4}\div \left ( -\frac{5}{7} \right )$.
a. $1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}$
$=1+\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}$
$=1+\frac{4}{23}-\frac{4}{23}+\frac{5}{21}+\frac{16}{21}+0,5$
$=(1+0,5)+\left (\frac{4}{23}-\frac{4}{23} \right )+\left (\frac{5}{21}+\frac{16}{21} \right )$
$=1,5+0+\frac{5+16}{21}$
$=1,5+1$
$=2,5$
b. $\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}$
$=\frac{3}{7}.\left (19\frac{1}{3}-33\frac{1}{3} \right )$
$=\frac{3}{7}.\left (19+\frac{1}{3}-33-\frac{1}{3} \right )$
$=\frac{3}{7}.\left (19-33+\frac{1}{3}-\frac{1}{3} \right )$
$=\frac{3}{7}.(-14)$
$=-6$
c. $9.\left ( -\frac{1}{3} \right )^{3}+\frac{1}{3}$
$=9.\left ( -\frac{1}{27} \right )+\frac{1}{3}$
$=-\frac{9}{27}+\frac{1}{3}$
$=-\frac{1}{3}+\frac{1}{3}$
$=0$
d. $15\frac{1}{4}\div \left ( -\frac{5}{7} \right )-25\frac{1}{4}\div \left ( -\frac{5}{7} \right )$
$=\left (15\frac{1}{4}-25\frac{1}{4} \right )\div \left ( -\frac{5}{7} \right )$
$=\left (15+\frac{1}{4}-25-\frac{1}{4} \right )\div \left ( -\frac{5}{7} \right )$
$=\left (15-25+\frac{1}{4}-\frac{1}{4} \right )\div \left ( -\frac{5}{7} \right )$
$=(-10).\left ( -\frac{7}{5} \right )$
$=14$
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