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a, \(4^3.5^3=\left(4.5\right)^3=20^3=8000\)
b, \(6^3.5^3=\left(6.5\right)^3=30^3=27000\)
c, \(8^2.5^2=\left(8.5\right)^2=40^2=1600\)
d, \(125^3.8^3=\left(125.8\right)^3=1000^3\)
e, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)
1. a) (x-2)2 =1
=> x - 2 = \(\pm\sqrt{1}\)
=> x - 2 = 1 hoặc -1
=> x = 3 hoặc 1
b) 2x - 1= -8
=> 2x = -7
=>x = \(\dfrac{-7}{2}\)
c)thiếu đề
d) (x-1)x+2 = (x-1)x+4
(x-1)x+2 = (x-1)x+2+2
(x-1)x+2 = (x-1)x+2. (x-1)2
(x-1)x+2 - (x-1)x+2. (x-1)2 = 0
=> (x-1)x+2. [1 - (x-1)2] = 0
\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)
b) \(\dfrac{7}{15}-\dfrac{9}{19}\)\(-\dfrac{-8}{15}-\dfrac{10}{19}\)
=\(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) \(-\left(\dfrac{9}{19}-\dfrac{10}{19}\right)\)
= \(-\dfrac{1}{15}\) - \(\left(-\dfrac{1}{19}\right)\)
\(=-\dfrac{1}{15}\) + \(\dfrac{1}{19}\)
= \(-\dfrac{4}{285}\)
c) \(1\dfrac{1}{3}\) \(\div\) \(\dfrac{4}{5}\) + 2\(\dfrac{2}{3}\) \(\div\)\(\dfrac{4}{5}\)
= \(\left(1\dfrac{1}{3}+2\dfrac{2}{3}\right)\) \(\div\dfrac{4}{5}\)
= \(\left[\left(1+2\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\) \(\div\dfrac{4}{5}\)
= ( 3 + 1 ) \(\div\dfrac{4}{5}\)
= 4 \(\div\dfrac{4}{5}\)
= \(\dfrac{4.5}{4}\)
= 5
Câu 1 :
\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)
\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)
\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)
Câu 2 :
\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)
Sorry . Mình chỉ biết đến đây thôi
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
a) \(\left(0,25\right)^3\cdot32=0,015625\cdot32=0,5\)
b) \(\left(-0,125\right)^3\cdot80^4=\dfrac{-1}{512}\cdot40960000=80000\)
c) \(\dfrac{8^2\cdot4^5}{2^{20}}=\dfrac{2^{3^2}\cdot2^{2^5}}{2^{20}}=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{2^{16}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\)
d) \(\dfrac{81^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{3^{4^{11}}\cdot3^{17}}{3^{3^{10}}\cdot3^{2^{15}}}=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{3^{61}}{3^{60}}=3\)
1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
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