Giải câu 1 trang 27 sách phát triển năng lực toán 7 tập 1

Giải câu 1 trang 27 sách phát triển năng lực toán 7 tập 1.

a. $(\frac{5}{7}-\frac{7}{5})-[\frac{1}{2}-(\frac{-2}{7}-\frac{1}{10})]$

= $(\frac{5}{7}-\frac{2}{7})-(\frac{7}{5}+\frac{1}{2}+\frac{1}{10})$

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

= $\frac{-11}{7}$

b. $\frac{-1}{4}.13\frac{9}{11}-0,25.6\frac{2}{11}$

= $-0,25.13\frac{9}{11}-0,25.6\frac{2}{11}$

= $-0,25(13\frac{9}{11}+6\frac{2}{11})$

= -0,25.20

= -5

c. $(\frac{-3}{4}+\frac{5}{13}):\frac{2}{7}-(2\frac{1}{4}-\frac{8}{13}):\frac{2}{7}$

= $[(\frac{-3}{4}+\frac{5}{13})-(2\frac{1}{4}-\frac{8}{13})]:\frac{2}{7}$

= $[(\frac{-3}{4}-2\frac{1}{4})+(\frac{5}{13}+\frac{8}{13})]:\frac{2}{7}$

= $(-3+1).\frac{7}{2}$

= $-2.\frac{7}{2}$

= -7

d. $\frac{3}{5}:(\frac{-1}{15}-\frac{1}{6})+\frac{3}{5}:(\frac{-1}{3}-1\frac{1}{15})$

= $\frac{3}{5}:(\frac{-2}{30}-\frac{5}{30})+\frac{3}{5}:(\frac{-5}{15}-\frac{-16}{15})$

= $\frac{3}{5}:\frac{-7}{30}+\frac{3}{5}:\frac{-7}{5}$

= $\frac{3}{5}.\frac{-30}{7}+\frac{3}{5}.\frac{-5}{7}$

= $\frac{3}{5}.(\frac{-30}{7}+\frac{-5}{7})$

= $\frac{3}{5}.(-5)$

= -3

e. B = $\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}$

     = $\frac{1.(\frac{1}{9}-\frac{1}{7}-\frac{1}{11})}{4.(\frac{1}{9}-\frac{1}{7}-\frac{1}{11})}+\frac{3.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}{4.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}$

     = $\frac{1}{4}+\frac{3}{4}$

     = 1

f. $2^3+3{\left(\frac{-1}{2}\right)}^0-{\left(\frac{1}{2}\right)}^2.4+[(-2{)}^2:\frac{1}{2}]:8$

 = $8+3.1-\frac{1}{2^2}.4+4.2:8$

 = 8 + 3 - 1 + 1

 = 11

g. $\frac{{81}^{11}{.3}^{17}}{{27}^{10}{.9}^{15}}$

  = $\frac{(3^4{)}^{11}{.3}^{17}}{(3^3{)}^{10}.(3^2{)}^{15}}$

  = $\frac{3^{44}{.3}^{17}}{3^{30}{.3}^{30}}$

  = $\frac{3^{61}}{3^{60}}$

  = $3^1$

  = 3

h. $\frac{4^6{.9}^5+6^9.120}{8^4{.3}^{12}-6^{11}}$

  = $\frac{(2^2{)}^6.(3^2{)}^5+(2.3{)}^9{.2}^3.3.5}{(2^3{)}^4{.3}^{12}-(2.3{)}^{11}}$

  = $\frac{2^{12}{.3}^{10}+2^{12}{.3}^{10}.5}{2^{12}{.3}^{12}-2^{11}{.3}^{11}}$

  = $\frac{2^{11}{.3}^{10}(2+2.5)}{2^{11}{.3}^{10}.({2.3}^2-3)}$

  = $\frac{12}{15}$

  = $\frac{4}{5}$