Câu 8: trang 10 - sgk toán 7 tập 1
Tính:
a. $\frac{3}{7}+(-\frac{5}{2})+(-\frac{3}{5})$
b. $(-\frac{4}{3}+(-\frac{2}{5})+(-\frac{3}{2})$
c. $\frac{4}{5}-(-\frac{2}{7})-\frac{7}{10}$
d. $\frac{2}{3}-\left [ (-\frac{7}{4})-(\frac{1}{2}+\frac{3}{8}) \right ]$
Bài Làm:
Ta có :
a. $\frac{3}{7}+(-\frac{5}{2})+(-\frac{3}{5})$
= $\frac{30}{70}+(-\frac{175}{70})+(-\frac{42}{70})$
= $\frac{-187}{70}=-2\frac{47}{70}$
Vậy $\frac{3}{7}+(-\frac{5}{2})+(-\frac{3}{5})=-2\frac{47}{70}$
b. $(-\frac{4}{3}+(-\frac{2}{5})+(-\frac{3}{2})$
= $(-\frac{40}{30}+(-\frac{12}{30})+(-\frac{45}{30})$
= $(-\frac{97}{30}=-3\frac{7}{30}$
Vậy $(-\frac{4}{3}+(-\frac{2}{5})+(-\frac{3}{2})=-3\frac{7}{30}$
c. $\frac{4}{5}-(-\frac{2}{7})-\frac{7}{10}$
= $\frac{56}{70}-(-\frac{20}{70})-\frac{49}{70}$
= $\frac{27}{70}$
Vậy $\frac{4}{5}-(-\frac{2}{7})-\frac{7}{10}=\frac{27}{70}$
d. $\frac{2}{3}-\left [ (-\frac{7}{4})-(\frac{1}{2}+\frac{3}{8}) \right ]$
= $\frac{2}{3}-\left [ (-\frac{7}{4})-(\frac{4}{8}+\frac{3}{8}) \right ]$
= $\frac{2}{3}-\left [ (-\frac{7}{4})-\frac{7}{8} \right ]$
= $\frac{2}{3}-\left [ (-\frac{14}{8})-\frac{7}{8} \right ]$
= $\frac{2}{3}-(-\frac{21}{8})$
= $\frac{16}{24}+\frac{63}{24}$
= $\frac{79}{24}$
Vậy $\frac{2}{3}-\left [ (-\frac{7}{4})-(\frac{1}{2}+\frac{3}{8}) \right ]=\frac{79}{24}$