Bài 4 trang 30 Toán 7 Tập 1 | Cánh diều Giải Toán lớp 7

Giải Toán lớp 7 Bài tập cuối chương 1

Bài 4 trang 30 Toán lớp 7 Tập 1: Tính:

a) $\mathrm{0,3}-\frac{4}{9}:\frac{4}{3}.\frac{6}{5}+1$;

b) ${\left(\frac{-1}{3}\right)}^2-\frac{3}{8}:{\left(\mathrm{0,5}\right)}^3-\frac{5}{2}.(-4)$;

c) $1+2:\left(\frac{2}{3}-\frac{1}{6}\right).(-\mathrm{2,25})$;

d) $\left.\left(\frac{1}{4}-\mathrm{0,5}\right).2+\frac{8}{3}\right]:2$.

Lời giải:

a)  $\mathrm{0,3}-\frac{4}{9}:\frac{4}{3}.\frac{6}{5}+1$

$=0,3-\frac{4}{9}.\frac{3}{4}.\frac{6}{5}+1$

$=0,3-\frac{\mathrm{4.3.6}}{\mathrm{9.4.5}}+1$

$=0,3-\frac{4.3.3.2}{3.3.4.5}+1$

$=0,3-\frac{2}{5}+1$

= 0,3 – 0,4 + 1

= 0,9.

b)  ${\left(\frac{-1}{3}\right)}^2-\frac{3}{8}:{\left(\mathrm{0,5}\right)}^3-\frac{5}{2}.(-4)$

$=\frac{1}{9}-\frac{3}{8}:{\left(\frac{1}{2}\right)}^3+\frac{5}{2}.4$

$=\frac{1}{9}-\frac{3}{8}:\frac{1}{8}+10$

$=\frac{1}{9}-\frac{3}{8}.\frac{8}{1}+10$

$=\frac{1}{9}-3+10$

$=\frac{1}{9}+(10-3)$

$=\frac{1}{9}+7$

$=\frac{1}{9}+\frac{63}{9}$

$=\frac{64}{9}$.

c)  $1+2:\left(\frac{2}{3}-\frac{1}{6}\right).(-\mathrm{2,25})$

$=1+2:\left(\frac{4}{6}-\frac{1}{6}\right).\frac{-9}{4}$

$=1+2:\frac{3}{6}.\frac{-9}{4}$

$=1+2:\frac{1}{2}.\frac{-9}{4}$

$=1+2.2.\frac{-9}{4}$

$=1+4.\frac{-9}{4}$

= 1 – 9

= – 8.

d)  $\left.\left(\frac{1}{4}-\mathrm{0,5}\right).2+\frac{8}{3}\right]:2$

$=\left.\left(\frac{1}{4}-\frac{1}{2}\right).2+\frac{8}{3}\right].\frac{1}{2}$

$=\left.\left(\frac{1}{4}-\frac{2}{4}\right).2+\frac{8}{3}\right].\frac{1}{2}$

$=\left.\frac{-1}{4}.2+\frac{8}{3}\right].\frac{1}{2}$

$=\left.\frac{-1}{2}+\frac{8}{3}\right].\frac{1}{2}$

$=\left.\frac{-3}{6}+\frac{16}{6}\right].\frac{1}{2}$

$=\frac{13}{6}.\frac{1}{2}$

$=\frac{13}{12}.$