Giải bài 1.8 trang 13

Đề bài

Tính giá trị của các biểu thức sau:

$\begin{array}{c}a\left)\right(8+2\frac{1}{3}-\frac{3}{5})-(5+0,4)-(3\frac{1}{3}-2)\\ b\left)\right(7-\frac{1}{2}-\frac{3}{4}):(5-\frac{1}{4}-\frac{5}{8})\end{array}$

Lời giải chi tiết

a) Cách 1:

$\begin{array}{c}(8+2\frac{1}{3}-\frac{3}{5})-(5+0,4)-(3\frac{1}{3}-2)\\ =(8+\frac{7}{3}-\frac{3}{5})-(5+\frac{4}{10})-(\frac{10}{3}-2)\\ =8+\frac{7}{3}-\frac{3}{5}-5-\frac{2}{5}-\frac{10}{3}+2\\ =(8-5+2)+(\frac{7}{3}-\frac{10}{3})-(\frac{3}{5}+\frac{2}{5})\\ =5+\frac{-3}{3}-\frac{5}{5}\\ =5+(-1)-1\\ =3\end{array}$

Cách 2:

$\begin{array}{c}(8+2\frac{1}{3}-\frac{3}{5})-(5+0,4)-(3\frac{1}{3}-2)\\ =(8+\frac{7}{3}-\frac{3}{5})-(5+\frac{4}{10})-(\frac{10}{3}-2)\\ =(\frac{120}{15}+\frac{35}{15}-\frac{9}{15})-(\frac{25}{5}+\frac{2}{5})-(\frac{10}{3}-\frac{6}{3})\\ =\frac{146}{15}-\frac{27}{5}-\frac{4}{3}\\ =\frac{146}{15}-\frac{81}{15}-\frac{20}{15}\\ =\frac{45}{15}\\ =3\end{array}$

b)

$\begin{array}{c}(7-\frac{1}{2}-\frac{3}{4}):(5-\frac{1}{4}-\frac{5}{8})\\ =(\frac{28}{4}-\frac{2}{4}-\frac{3}{4}):(\frac{40}{8}-\frac{2}{8}-\frac{5}{8})\\ =\frac{23}{4}:\frac{33}{8}\\ =\frac{23}{4}.\frac{8}{33}\\ =\frac{46}{33}\end{array}$