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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
1, Tìm x biết: a, 6x 1-6x=1080
b, 6x-1 6x=42 2, So sánh: E=7.(8 82 83 ....... 8100) 8 và G=8101 3, Chứng tỏ: a, 4343-1717 chia hết cho 10 b, 3636-910 chia hết cho 45
c, 2 10 2 11 2 12 7 210 211 2127 có giá trị là số tự nhiên
d, 8 10 − 8 9 − 8 8 55 810−89−8855 có giá trị là số tự nhiên
hi 
Bài 1:
a: \(\Leftrightarrow6^x\left(6-1\right)=1080\)
=>6x=216
=>x=3
b: \(\Leftrightarrow6^x\left(\dfrac{1}{6}+1\right)=42\)
=>6x=36
=>x=2
Câu 3:
c: \(=\dfrac{2^{10}\left(1+2+2^2\right)}{7}=2^{10}\) là số tự nhiên
d: \(=\dfrac{8^8\left(8^2-8-1\right)}{55}=8^8\) là số tự nhiên
\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a) \(10^{20}\) và \(9^{10}\)
Vì 10 > 9 ; 20 > 10
nên \(10^{20}>9^{10}\)
Vậy \(10^{20}>9^{10}\)
b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)
Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
Vì 243 > 125 nên \(125^{10}< 243^{10}\)
Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) \(64^8\) và \(16^{12}\)
Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Vậy \(64^8=16^{12}\left(=4^{24}\right)\)
d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
a) 814=(23)14=23*14=242
1610=(8*2)10=810*210=(23)10*210=230*210=240
Vì 242 > 240 nên 814 > 1610
b) 233=(23)11=811
322=(32)11=911
Vì 811 < 911 nên 233 < 322
Bài 1: So Sánh
a) Ta có: \(2^{100}=2^{10^{10}}=1024^{10}\)
\(10^{30}=10^{3\cdot10}=1000^{10}\)
mà \(1024^{10}>1000^{10}\)
nên \(2^{100}>10^3\)
b) Ta có: \(5\cdot8^{25}=5\cdot2^{75}\)
\(128^{11}=2^{77}=4\cdot2^{75}\)
mà \(5\cdot2^{75}>4\cdot2^{75}\)
nên \(5\cdot8^{25}>128^{11}\)
c) Ta có: \(8\cdot27^6=8\cdot3^{18}\)
\(9^{10}=3^{20}=9\cdot3^{18}\)
mà \(8\cdot3^{18}< 9\cdot3^{18}\)
nên \(8\cdot27^6< 9^{10}\)
d) Ta có: \(2^{100}=2^{69}\cdot2^{31}\)
\(=2^{31}\cdot2^{63}\cdot2^6\)
\(=2^{31}\cdot\left(2^9\right)^7\cdot\left(2^2\right)^3\)
\(=2^{31}\cdot512^7\cdot4^3\)
Ta có: \(10^{31}=2^{31}\cdot5^{31}\)
\(=2^{31}\cdot5^{28}\cdot5^3\)
\(=2^{31}\cdot\left(5^4\right)^7\cdot5^3\)
\(=2^{31}\cdot625^7\cdot5^3\)
Ta có: \(512^7< 625^7\)
\(4^3< 5^3\)
Do đó: \(512^7\cdot4^3< 625^7\cdot5^3\)
\(\Leftrightarrow2^{31}\cdot512^7\cdot4^3< 2^{31}\cdot625^7\cdot5^3\)
hay \(2^{100}< 10^{31}\)
\(a,\left(-0,125\right)^4=0.125^4=\left(0.5^3\right)^4=0.5^{12}\)
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
d) 812 và 128
Ta có : 812 = (96)2 = 5314412
128 = (124)2 = 207362
=> 534412 > 207362
=> 812 > 128
a) 36 và 63
Ta có : 36 = (32)3 = 93
63 = 63
=> 93 > 63
=> 36 > 63