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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

A= 4/7.

Biết có cái

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

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

-_-

a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)

c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)

d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)

\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)

\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)

e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)

\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)

\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)

f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)

\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)

\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)

g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)

\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)

\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)

\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)

\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)

\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)

\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)

\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)

\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)

\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)

\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)

\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)

\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)

\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)

Vậy \(P=\dfrac{37}{60}\)

\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)

\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)

\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)

\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)

Vậy \(Q=\dfrac{29}{125}\)

a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)

b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)

c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)

\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)

d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)

\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)

e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)

\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)

\(=\left(3-1\right)^{106}=2^{106}\)

f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)

g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)

\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)

\(=2023^{4\sqrt{2}}\)

\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{125}{100}+\dfrac{79}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{5}{4}+\dfrac{79}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)

\(=-5,13:\dfrac{57}{14}=-5,13:\dfrac{15}{57}\)

\(=\dfrac{-71,82}{57}=1,26\)

Vậy \(A=1,26\)

\(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)

\(=\left(\dfrac{10}{3}.1,9+19,5:\dfrac{13}{3}\right).\left(\dfrac{62-12}{75}\right)\)

\(=\left(\dfrac{19}{3}+\dfrac{58,5}{13}\right).\dfrac{50}{75}\)

\(=\left(\dfrac{19}{3}+4,5\right).\dfrac{2}{3}\)

\(=\dfrac{32,5}{3}.\dfrac{2}{3}=\dfrac{65}{9}=7\dfrac{2}{9}\)

Vậy \(B=7\dfrac{2}{9}\)

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.=\(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{61}{4}:\dfrac{7}{12}=\dfrac{471}{14}-\dfrac{183}{7}=\dfrac{15}{2}\)

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

c.\(\left(\dfrac{10}{3}+2.5\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}=\dfrac{35}{6}:\left(-\dfrac{31}{30}\right)-\dfrac{11}{31}=-\dfrac{175}{31}-\dfrac{11}{31}=-6\)

d.\(\left[6+\dfrac{1}{8}-\dfrac{1}{2}\right]:\dfrac{3}{12}=\dfrac{45}{8}:\dfrac{3}{12}=\dfrac{45}{2}\)

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

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