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a) \(\frac{2^2.9^2}{6^4.8}\)\(=\frac{2^2+\left(3^2\right)^2}{\left(2.3\right)^4.2^3}\)\(=\frac{3^4}{2^4.3^4.2}=\frac{1}{2^4.2}=\frac{1}{2^5}=\frac{1}{32}\)
b)\(\frac{3^{10}.2^1}{16.4^3.243}=\frac{3^{10}.2^1}{2^4.4^3.3^5}=\frac{3^5}{2^3.4^3}=\frac{3^5}{\left(2.4\right)^3}\)\(=\frac{3^5}{8^3}=\frac{243}{512}\)
1:a)\(\frac{28^{15}\cdot3^{17}}{84^{16}}\)=\(\frac{28^{15}\cdot3^{15}\cdot3^2}{84^{16}}\)=\(\frac{\left(28^{15}\cdot3^{15}\right)\cdot3^2}{84^{16}}\)=\(\frac{84^{15}\cdot9}{84^{16}}\)=\(\frac{9}{84}\)=\(\frac{3}{28}\)
b)\(\frac{3^{10}+6^2}{5\cdot3^8+20}\)=\(\frac{3^2\cdot3^8+2^2\cdot3^2}{5\cdot3^8+5\cdot4}\)=\(\frac{9\cdot3^8+4\cdot9}{5\cdot\left(3^8+4\right)}\)=\(\frac{9\cdot\left(3^8+4\right)}{5\cdot\left(3^8+4\right)}\)=\(\frac{9}{5}\)
Xét vế trái ta có:72^15=3^15*24^15=3^15*24^9*24^6
=3^15*24^9*12^6*2^6=3^15*24^9*12^6*4^3(1)
Xét vế phải ta có: 3^21*96^9=3^15*3^6*24^9*4^9
=3^15*3^6*24^9*4^6*4^3=3^15*24^9*(3^6*4^6*4^3)
=3^15*24^9*(12^6*4^3)(2)
từ (1) và (2)=>72^15=3^21*96^9
Bây giờ mình sẽ viết đầy đủ hơn nhé:
a) \(\frac{28^{15}.3^{17}}{84^{16}}=\frac{\left(28.3\right)^{15}.3^2}{\left(28.3\right)^{15}.84}=\frac{9}{84}=\frac{3}{28}\)
b)\(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+2^2.3^2}{5.3^8+20}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}\)\(=\frac{9}{5}\)
a) \(\frac{28^{15}.3^{17}}{84^{16}}=\frac{\left(2^2.7\right)^{15}.3^{17}}{\left(2^2.3.7\right)^{16}}=\frac{2^{30}.7^{15}.3^{17}}{2^{32}.3^{16}.7^{16}}=\frac{3}{2^2.7}=\frac{3}{28}\)
b) \(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+\left(2.3\right)^2}{5.3^9+2^2.5}=\frac{3^{10}+2^2.3^2}{5\left(3^8.2^2\right)}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}=\frac{3^2}{5}=\frac{9}{5}\)
\(\dfrac{x-1}{50}+\dfrac{x-2}{49}=\dfrac{x-3}{48}+\dfrac{x-4}{47}\)
\(\Rightarrow\dfrac{x-1}{50}-1+\dfrac{x-2}{49}-1=\dfrac{x-3}{48}-1+\dfrac{x-4}{47}-1\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}=\dfrac{x-51}{48}+\dfrac{x-51}{47}\)
\(\Rightarrow\dfrac{x-51}{50}+\dfrac{x-51}{49}-\dfrac{x-51}{48}-\dfrac{x-51}{47}=0\)
\(\Rightarrow\left(x-51\right)\left(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\right)=0\)
Vì \(\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{48}-\dfrac{1}{47}\ne0\) nên \(x-51=0\Rightarrow x=51\)
\(\dfrac{x+25}{6}+\dfrac{x+20}{11}+\dfrac{x+16}{15}+3=0\)
\(\Rightarrow\dfrac{x+25}{6}+1+\dfrac{x+20}{11}+1+\dfrac{x+16}{15}+1=0\)
\(\Rightarrow\dfrac{x+31}{6}+\dfrac{x+31}{11}+\dfrac{x+31}{15}=0\)
\(\Rightarrow\left(x+31\right)\left(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}+\dfrac{1}{15}\ne0\) nên \(x+31=0\Rightarrow x=-31\)
\(\dfrac{x-15}{6}+\dfrac{x-10}{11}=\dfrac{x-3}{18}+\dfrac{x-7}{14}\)
\(\Rightarrow\dfrac{x-15}{6}-1+\dfrac{x-10}{11}-1=\dfrac{x-3}{18}-1+\dfrac{x-7}{14}-1\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}=\dfrac{x-21}{18}+\dfrac{x-21}{14}\)
\(\Rightarrow\dfrac{x-21}{6}+\dfrac{x-21}{11}-\dfrac{x-21}{18}-\dfrac{x-21}{14}=0\)
\(\Rightarrow\left(x-21\right)\left(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{6}+\dfrac{1}{11}-\dfrac{1}{18}-\dfrac{1}{14}\ne0\) nên \(x-21=0\Rightarrow x=21\)
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)
b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)
e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)
=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)
=\(-\dfrac{516}{7}\)
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)
=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)
=\(\dfrac{119}{240}\)
a) \(\dfrac{15^{30}}{45^{15}}=\dfrac{15^{30}}{3^{15}.15^{15}}=\dfrac{15^{15}}{3^{15}}=5^{15}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{8^5.3^8}{2^6.3^6.8^3}=\dfrac{8^2.3^2}{2^6}=\dfrac{2^6.3^2}{2^6}=3^2=9\)
c) \(\dfrac{14^{10}.21^{32}.35^{48}}{10^{10}.15^{32}.7^{96}}=\dfrac{2^{10}.7^{10}.3^{32}.7^{32}.5^{48}.7^{48}}{2^{10}.5^{10}.3^{32}.5^{32}.7^{96}}\)
= \(\dfrac{2^{10}.7^{58}.3^{32}.5^{48}}{2^{10}.5^{42}.3^{32}.7^{96}}=\dfrac{5^6}{7^{38}}\) ( Câu này làm bừa, có lẽ sai đấy :)) )
2. So sánh
a) 3200 = 9100
2300 = 8100
Vì 9100 > 8100 nên 3200 < 2300
b) 912 = 7294
268 = 6764
Vì 7294 > 6764 nên 912 > 268
c) 224 = 88
316 = 98
Vì 88 < 98 nên 224 < 316
a: \(A=\dfrac{4^{15}\cdot7^{15}\cdot3^{17}}{4^{32}\cdot3^{16}}=\dfrac{1}{4^{17}}\cdot3\cdot7^{15}=\dfrac{3\cdot7^{15}}{4^{17}}\)
b: \(B=\dfrac{3^{10}+6^2}{5\cdot3^8+20}=\dfrac{3^{10}+3^2\cdot4}{5\left(3^8+4\right)}=\dfrac{3^2\left(3^8+4\right)}{5\left(3^8+4\right)}=\dfrac{9}{5}\)