Giải bài 96 trang 48 - SGK Toán lớp 7 Tập 1
Thực hiện phép tính (bằng cách hợp lí nếu có thể):
a) \(1\dfrac{4}{23} + \dfrac{5}{21} - \dfrac{4}{23} + 0,5 + \dfrac{16}{21};\)
b) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}. 33\dfrac{1}{3};\)
c) \(9.\left(-\dfrac{1}{3}\right)^3 + \dfrac{1}{3};\)
d) \(15\dfrac{1}{4}\ : \left(-\dfrac{5}{7}\right) - 25 \dfrac{1}{4} : \left(-\dfrac{5}{7}\right);\)
a) \(1\dfrac{4}{23} + \dfrac{5}{21} - \dfrac{4}{23} + 0,5 + \dfrac{16}{21}\)
\(= 1+ \dfrac{4}{23} + \dfrac{5}{21} - \dfrac{4}{23} + 0,5 + \dfrac{16}{21}\)
\(=1 + 0,5 + 1\)
\(= 2,5\)
b) \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}. 33\dfrac{1}{3}\)
\(= \dfrac{3}{7}.\left(19\dfrac{1}{3}- 33\dfrac{1}{3}\right)\)
\( = \dfrac{3}{7}.\left(19 + \dfrac{1}{3}- 33 -\dfrac{1}{3}\right)\)
\(= \dfrac{3}{7}.(-14)\)
\(= \dfrac{-42}{7} = -6\)
c) \(9.\left(-\dfrac{1}{3}\right)^3 + \dfrac{1}{3}\)
\(= 9.\left(-\dfrac{1}{27}\right) + \dfrac{1}{3}\)
\(= -\dfrac{1}{3} + \dfrac{1}{3}\)
\(= 0\)
d) \(15\dfrac{1}{4}\ : \left(-\dfrac{5}{7}\right) - 25 \dfrac{1}{4} : \left(-\dfrac{5}{7}\right)\)
\(= \left(15\dfrac{1}{4}\ -25\dfrac{1}{4}\right) : \left(-\dfrac{5}{7}\right)\)
\(= \left(15 + \dfrac{1}{4}\ -25-\dfrac{1}{4}\right) : \left(-\dfrac{5}{7}\right)\)
\(= -10 : \left(-\dfrac{5}{7}\right)\)
\(= -10. \left(-\dfrac{7}{5}\right)\)
\(= \dfrac{70}{5} = 14\)
