Giải câu 1 trang 27 sách phát triển năng lực toán 7 tập 1.
a. $\left ( \frac{5}{7}-\frac{7}{5} \right )-\left [ \frac{1}{2}-\left ( \frac{-2}{7}-\frac{1}{10} \right ) \right ]$
= $\left ( \frac{5}{7}-\frac{2}{7} \right ) - \left ( \frac{7}{5}+\frac{1}{2}+\frac{1}{10} \right )$
= $\frac{3}{7} - 2$
= $\frac{-11}{7}$
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. $\left ( \frac{-3}{4}+\frac{5}{13} \right ):\frac{2}{7}-\left ( 2\frac{1}{4}-\frac{8}{13} \right ):\frac{2}{7}$
= $\left [ \left (\frac{-3}{4}+\frac{5}{13} \right )-\left ( 2\frac{1}{4}-\frac{8}{13} \right ) \right ]:\frac{2}{7}$
= $\left [ \left ( \frac{-3}{4}-2\frac{1}{4} \right )+\left ( \frac{5}{13}+\frac{8}{13} \right ) \right ]:\frac{2}{7}$
= $(-3+1).\frac{7}{2}$
= $-2.\frac{7}{2}$
= -7
d. $\frac{3}{5}:\left ( \frac{-1}{15}-\frac{1}{6} \right )+\frac{3}{5}:\left ( \frac{-1}{3}-1\frac{1}{15} \right )$
= $\frac{3}{5}:\left (\frac{-2}{30}-\frac{5}{30} \right )+\frac{3}{5}:\left ( \frac{-5}{15}-\frac{-16}{15} \right )$
= $\frac{3}{5}:\frac{-7}{30}+\frac{3}{5}:\frac{-7}{5}$
= $\frac{3}{5}.\frac{-30}{7}+\frac{3}{5}.\frac{-5}{7}$
= $\frac{3}{5}.\left (\frac{-30}{7}+\frac{-5}{7} \right )$
= $\frac{3}{5}.(-5)$
= -3
e. B = $\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}$
= $\frac{1.\left (\frac{1}{9}-\frac{1}{7}-\frac{1}{11} \right )}{4.\left (\frac{1}{9}-\frac{1}{7}-\frac{1}{11} \right )}+\frac{3.\left ( \frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625} \right )}{4.\left ( \frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625} \right )}$
= $\frac{1}{4}+\frac{3}{4}$
= 1
f. $2^{3}+3\left ( \frac{-1}{2} \right )^{0}-\left ( \frac{1}{2} \right )^{2}.4+\left [ (-2)^{2}:\frac{1}{2} \right ]:8$
= $8 + 3.1 - \frac{1}{2^{2}}.4 + 4.2:8$
= 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. $\frac{4^{6}.9^{5}+6^{9}.120}{8^{4}.3^{12}-6^{11}}$
= $\frac{(2^{2})^{6}.(3^{2})^{5}+(2.3)^{9}.2^{3}.3.5}{(2^{3})^{4}.3^{12}-(2.3)^{11}}$
= $\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}$
= $\frac{2^{11}.3^{10}(2+2.5)}{2^{11}.3^{10}.(2.3^{2}-3)}$
= $\frac{12}{15}$
= $\frac{4}{5}$