Giải SBT Toán 7 Bài 4. Quy tắc dấu ngoặc và quy tắc chuyển vế có đáp án

a) $(\frac{-3}{8})+(\frac{7}{9}-\frac{5}{8})=(\frac{-3}{8})+\frac{7}{9}-\frac{5}{8}$

$=(\frac{-3}{8})-\frac{5}{8}+\frac{7}{9}=-1+\frac{7}{9}$

$=\frac{-9}{9}+\frac{7}{9}=\frac{-2}{9}$

b) $[(\frac{-7}{8}):\frac{21}{16}]-\frac{5}{3}.(\frac{1}{3}-\frac{7}{10})$

$=[(\frac{-7}{8}).\frac{16}{21}]-\frac{5}{3}.(\frac{10}{30}-\frac{21}{30})$

$=\frac{(-7)\ .\ 16}{8\ .\ 21}-\frac{5}{3}.(\frac{-11}{30})$

$=-\frac{7\ .\ 8\ .\ 2}{8\ .\ 7\ .\ 3}+\frac{5\ .\ 11}{3\ .\ 5\ .\ 6}$

$=\frac{-2}{3}+\frac{11}{18}=\frac{-1}{18}$

c) ${[(\frac{-2}{3})+\frac{3}{4}]}^{2}.\frac{12}{5}-\frac{1}{5}$

$={[(\frac{-8}{12})+\frac{9}{12}]}^{2}.\frac{12}{5}-\frac{1}{5}$

$={(\frac{1}{12})}^{2}.\frac{12}{5}-\frac{1}{5}$

$=\frac{1}{{12}^{2}}.\frac{12}{5}-\frac{1}{5}$

$=\frac{1}{60}-\frac{1}{5}=\frac{-11}{60}$

a) $(-0{,}5)-(-1+\frac{2}{3}):1{,}5+(\frac{-1}{4})$

$=(\frac{-1}{2})-(\frac{-3}{3}+\frac{2}{3}):\frac{3}{2}+(\frac{-1}{4})$

$=(\frac{-1}{2})-(\frac{-1}{3}).\frac{2}{3}+(\frac{-1}{4})$

$=(\frac{-1}{2})+\frac{2}{9}+(\frac{-1}{4})$

$=(\frac{-18}{36})+\frac{8}{36}+(\frac{-9}{36})=\frac{-19}{36}$

b) $\frac{4}{9}-(\frac{3}{7}+\frac{2}{9})=\frac{4}{9}-\frac{3}{7}-\frac{2}{9}$

$=\frac{4}{9}-\frac{2}{9}-\frac{3}{7}=\frac{2}{9}-\frac{3}{7}$

$=\frac{14}{63}-\frac{27}{63}=\frac{-13}{63}$

c) $[(\frac{-2}{5})+\frac{1}{3}]-(\frac{3}{5}-\frac{1}{4})=(\frac{-2}{5})+\frac{1}{3}-\frac{3}{5}+\frac{1}{4}$

$=(\frac{-2}{5})-\frac{3}{5}+\frac{1}{3}+\frac{1}{4}=(-1)+\frac{1}{3}+\frac{1}{4}$

$=(\frac{-12}{12})+\frac{4}{12}+\frac{3}{12}=\frac{-5}{12}$

d) $(1\frac{1}{2}-\frac{3}{4})-(0{,}25+\frac{1}{2})=(\frac{3}{2}-\frac{3}{4})-(\frac{1}{4}+\frac{1}{2})$

$=(\frac{3}{2}-\frac{3}{4})-(\frac{1}{4}+\frac{1}{2})$

$=(\frac{3}{2}-\frac{1}{2})-(\frac{3}{4}+\frac{1}{4})$

= 1 – 1 = 0.

a) $A=(8-\frac{2}{3}+\frac{1}{2})-(5-\frac{7}{3}-\frac{3}{2})-(\frac{5}{3}+\frac{5}{2}+4)$

$=(\frac{48}{6}-\frac{4}{6}+\frac{3}{6})-(\frac{30}{6}-\frac{14}{6}-\frac{9}{6})-(\frac{10}{6}+\frac{15}{6}+\frac{24}{6})$

$=\frac{47}{6}-\frac{7}{6}-\frac{49}{6}=\frac{-9}{6}=\frac{-3}{2}$

b) $A=(8-\frac{2}{3}+\frac{1}{2})-(5-\frac{7}{3}-\frac{3}{2})-(\frac{5}{3}+\frac{5}{2}+4)$

$=8-\frac{2}{3}+\frac{1}{2}-5+\frac{7}{3}+\frac{3}{2}-\frac{5}{3}-\frac{5}{2}-4$

$=(8-5-4)+(\frac{7}{3}-\frac{2}{3}-\frac{5}{3})+(\frac{1}{2}+\frac{3}{2}-\frac{5}{2})$

$=(-1)+0-\frac{1}{2}=\frac{-3}{2}$

d) ${(\frac{1}{25}-0{,}4)}^{2}:\frac{9}{125}-[(1\frac{1}{3}-\frac{2}{5}).\frac{3}{7}]$

$={(\frac{1}{25}-\frac{2}{5})}^{2}.\frac{125}{9}-[(\frac{4}{3}-\frac{2}{5}).\frac{3}{7}]$

$={(\frac{-9}{25})}^{2}.\frac{125}{9}-[\frac{14}{15}.\frac{3}{7}]$

$=\frac{{9}^{2}}{{25}^{2}}.\frac{125}{9}-\frac{2}{5}=\frac{{({3}^{2})}^{2}}{{({5}^{2})}^{2}}.\frac{{5}^{3}}{{3}^{2}}-\frac{2}{5}$

$=\frac{{3}^{4}}{{5}^{4}}.\frac{{5}^{3}}{{3}^{2}}-\frac{2}{5}=\frac{{3}^{2}}{5}-\frac{2}{5}$

$=\frac{9}{5}-\frac{2}{5}=\frac{7}{5}$

a) $x+\frac{3}{7}=\frac{2}{5}$

$x=\frac{2}{5}-\frac{3}{7}$

$x=\frac{14}{35}-\frac{15}{35}$

$x=\frac{-1}{35}$

b) $\frac{3}{2}-x=\frac{4}{5}$

$x=\frac{3}{2}-\frac{4}{5}$

$x=\frac{15}{10}-\frac{8}{10}$

$x=\frac{7}{10}$

c) $\frac{5}{9}-\frac{1}{3}x=\frac{2}{3}$

$\frac{1}{3}x=\frac{5}{9}-\frac{2}{3}$

$\frac{1}{3}x=\frac{-1}{9}$

$x=(\frac{-1}{9}):\frac{1}{3}$

$x=\frac{-1}{3}$

d) $\frac{3}{5}x-1\frac{1}{5}=\frac{-3}{14}:\frac{5}{7}$

$\frac{3}{55}x-\frac{6}{5}=\frac{-3}{10}$

$\frac{3}{5}x=\frac{-3}{10}+\frac{6}{5}$

$\frac{3}{5}x=\frac{9}{10}$

$x=\frac{3}{2}$

a) $\frac{3}{4}+\frac{1}{9}:x=0{,}5$

$\frac{1}{9}:x=\frac{1}{2}-\frac{3}{4}$

$\frac{1}{9}:x=\frac{-1}{4}$

$x=\frac{1}{9}:(\frac{-1}{4})$

$x=\frac{1}{9}\ .\ (-4)$

$x=\frac{-4}{9}\$

b) $\frac{3}{4}-(x-\frac{2}{3})=1\frac{1}{3}$

$x-\frac{2}{3}=\frac{3}{4}-\frac{4}{3}$

$x-\frac{2}{3}=\frac{-7}{12}$

$x=\frac{-7}{12}+\frac{2}{3}$

$x=\frac{1}{12}$

c) $(\frac{5}{7}-x).\frac{11}{15}=\frac{-22}{45}$

$\frac{5}{7}-x=(\frac{-22}{45}):\frac{11}{15}$

$\frac{5}{7}-x=(\frac{-22}{45}).\frac{15}{11}$

$\frac{5}{7}-x=\frac{-2}{3}$

$x=\frac{5}{7}-(\frac{-2}{3})$

$x=\frac{29}{21}$

d) $(2{,}5x-\frac{4}{7}):\frac{8}{21}=-1{,}5$

$(\frac{5}{2}x-\frac{4}{7}):\frac{8}{21}=\frac{-3}{2}$

$\frac{5}{2}x-\frac{4}{7}=(\frac{-3}{2}).\frac{8}{21}$

$\frac{5}{2}x-\frac{4}{7}=\frac{-4}{7}$

$\frac{5}{2}x=(\frac{-4}{7})+\frac{4}{7}$

$\frac{5}{2}x=0$

x=0

a) $\frac{12}{23}.\frac{7}{13}+\frac{11}{23}.\frac{7}{13}$

$=\frac{7}{13}\ .\ (\frac{12}{23}+\frac{11}{23})$

$=\frac{7}{13}\ .\ 1=\frac{7}{13}$

b) $\frac{4}{9}.\frac{23}{11}-\frac{1}{11}.\frac{4}{9}+\frac{4}{9}$

$=\frac{4}{9}.(\frac{23}{11}-\frac{1}{11}+1)$

$=\frac{4}{9}.(2+1)=\frac{4}{9}.3=\frac{4}{3}$

c) $[(-\frac{5}{7})+\frac{3}{5}]:\frac{2020}{2021}+(\frac{2}{5}-\frac{2}{7}):\frac{2020}{2021}$

$=[(-\frac{5}{7})+\frac{3}{5}].\frac{2021}{2020}+(\frac{2}{5}-\frac{2}{7}).\frac{2021}{2020}$

$=[(-\frac{5}{7})+\frac{3}{5}+\frac{2}{5}-\frac{2}{7}].\frac{2021}{2020}$

$=(1-1).\frac{2021}{2020}=0.\frac{2021}{2020}=0$

d) $\frac{3}{8}:(\frac{7}{22}-\frac{2}{11})+\frac{3}{8}:(\frac{2}{5}-\frac{1}{10})$

$=\frac{3}{8}:(\frac{7}{22}-\frac{4}{22})+\frac{3}{8}:(\frac{4}{10}-\frac{1}{10})$

$=\frac{3}{8}:\frac{3}{22}+\frac{3}{8}:\frac{3}{10}$

$=\frac{3}{8}.(\frac{22}{3}+\frac{10}{3})=\frac{3}{8}.\frac{32}{3}=4$