A= 4/7.

Biết có cái

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

a) $A=\frac{3}{5}.\frac{6}{7}+\frac{3}{7}.\frac{3}{5}-\frac{2}{7}.\frac{3}{5}$
$=\frac{3}{5}\cdot \left(\frac{6}{7}+\frac{3}{7}-\frac{2}{7}\right)=\frac{3}{5}$
b) $B=\left(-13\cdot \frac{2}{5}+\frac{-2}{9}\cdot \frac{2}{5}+\frac{2}{5}\cdot \frac{11}{9}\right)\cdot \frac{5}{2}$
$=\left(-13-\frac{2}{9}+\frac{11}{9}\right)\cdot \frac{2}{5}\cdot \frac{5}{2}=-13+\left(\frac{11}{9}-\frac{2}{9}\right)=-12.$
c) $C=\left(\frac{-4}{5}+\frac{5}{7}\right)\cdot \frac{3}{2}+\left(\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\frac{-4}{5}+\frac{5}{7}+\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\left(\frac{-4}{5}+\frac{-1}{5}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\right)\cdot \frac{3}{2}=0.$
d) $D=\frac{4}{9}:\left(\frac{1}{15}-\frac{10}{15}\right)+\frac{4}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)$
$=\frac{4}{9}:\frac{-3}{5}+\frac{4}{9}:\frac{-3}{22}=\frac{4}{9}\cdot \frac{-5}{3}+\frac{4}{9}.\frac{-22}{3}$
$=\frac{4}{9}\cdot \left(\frac{-5}{3}+\frac{-22}{3}\right)=\frac{4}{9}.\frac{-27}{3}=-4.$

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}.\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{3}{5}$

b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{5}{2}$
$=\left(-13-\genfrac{}{}{0ex}{}{2}{9}+\genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{5}{2}=-13+\left(\genfrac{}{}{0ex}{}{11}{9}-\genfrac{}{}{0ex}{}{2}{9}\right)=-12.$
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{2}{7}\right)\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=0.$
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{10}{15}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{2}{22}-\genfrac{}{}{0ex}{}{5}{22}\right)$

a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)

b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729

b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)

c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)

\(=\dfrac{2^3}{4^4}\cdot7\)

\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)

d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)

\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)

\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)

e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)

\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)

\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)

f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)

\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)

g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)

\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)

\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)

\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)

-_-

a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)

\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)

b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)

\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)

\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)

\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)

\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)

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