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a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)
b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)
c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)
\(B=\frac{1}{3}.b+\frac{2}{9}.b– b: \frac{9}{4}= \frac{1}{3}.b+\frac{2}{9}.b – b. \frac{4}{9}\)
\(=b(\frac{1}{3}+\frac{2}{9}-\frac{4}{9})=b. (\frac{3}{9}+\frac{2}{9}-\frac{4}{9})= b. \frac{1}{9} \)
Thay \(b=\frac{9}{10}\) vào B, ta được
B= \(b=\frac{9}{10}. \frac{1}{9}= \frac{1}{10}\)
a, =\(3^4+2^5=81+32=113\)
b, =\(3.\left(4^2-2.3\right)=3.\left(16-6\right)=3.10=30\)
c, =\(\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\dfrac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
d, =\(\dfrac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=3\)
e, =\(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}=\dfrac{3^{10}.2^{10}.5^7}{2^{10}.3^{10}.5^5}=5^2=25\)
g, =\(\dfrac{2^5.\left(2^8+1\right)}{2^2.\left(2^8+1\right)}=\dfrac{2^5}{2^2}=2^3=8\)
a, A = \(\dfrac{3^{10}\times10+3^{10}\times6}{3^9\times2^4}\)
A = \(\dfrac{3^{10}\times\left(10+6\right)}{3^9\times2^4}\)
A = \(\dfrac{3^{10}\times16}{3^9\times16}\)
A = 3
c, C = \(\dfrac{36^{10}\times25^{15}}{30^8}\)
C = \(\dfrac{\left(6^2\right)^{10}.\left(5^2\right)^{15}}{30^8}\)
C = \(\dfrac{6^{20}.5^{30}}{6^8.5^8}\)
C = 612.522
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
a) Ta có: \(\dfrac{39}{-65}=\dfrac{-39}{65}=\dfrac{-39:13}{65:13}=\dfrac{-3}{5}\)
\(\dfrac{-3}{5}=\dfrac{-3}{5}\)
Do đó: \(\dfrac{-3}{5}=\dfrac{39}{-65}\)
b) Ta có: \(\dfrac{-9}{27}=\dfrac{-9:9}{27:9}=\dfrac{-1}{3}\)
\(\dfrac{-41}{123}=\dfrac{-41:41}{123:41}=\dfrac{-1}{3}\)
Do đó: \(\dfrac{-9}{27}=\dfrac{-41}{123}\)
c) Ta có: \(\dfrac{-3}{4}=\dfrac{-3\cdot5}{4\cdot5}=\dfrac{-15}{20}\)
\(\dfrac{4}{-5}=\dfrac{-4}{5}=\dfrac{-4\cdot4}{5\cdot4}=\dfrac{-16}{20}\)
mà \(\dfrac{-15}{20}>\dfrac{-16}{20}\)
nên \(\dfrac{-3}{4}>\dfrac{4}{-5}\)
d) Ta có: \(\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot7}{3\cdot7}=\dfrac{-14}{21}\)
\(\dfrac{-5}{7}=\dfrac{-5\cdot3}{7\cdot3}=\dfrac{-15}{21}\)
mà \(\dfrac{-14}{21}>\dfrac{-15}{21}\)
nên \(\dfrac{2}{-3}>\dfrac{-5}{7}\)