Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ => x3 = 64 => x3 = 43 => x = 4
b/ => x5 = 32 => x5 = 25 => x = 2
c/ => 3x = 81 => 3x = 34 => x = 4
d/ => 3x+2 = 243 => 3x+2 = 35 => x + 2 = 5 => x = 3
e/ => (x - 2)3 = 64 => (x - 2)3 = 43 => x - 2 = 4 => x = 6
f/ => (x - 8)2 = 81 => (x - 8)2 = 92 => x - 8 = 9 => x = 17
7x-x=521:519+3.22-70
=> 6x = 5^2 + 12 -1
=> 6x = 36
=> x = 36/6 = 6
Kết quả 6
Học tốt
a) \(5^x.5^2=3^2+4^2\Leftrightarrow5^x.5^2=25\)
\(\Leftrightarrow5^x=\frac{25}{5^2}=1=5^0\Rightarrow x=0\)
b) \(\left(x-2\right)^3=27\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)
c) \(2018^{x-9}=1\). Ta có: \(2018^0=1\) nên để \(2018^{x-9}=1\) thì \(x-9=0\Leftrightarrow x=9\)
a) \(\frac{18^4.3^2.8^3}{27^3.16^2}=\frac{\left(2.3^2\right)^4.3^2.\left(2^3\right)^3}{\left(3^3\right)^3.\left(2^4\right)^2}=\frac{2^4.2^9.3^8.3^2}{3^9.2^8}=\frac{2^{13}.3^{10}}{3^9.2^8}=3.2^5=96\)
b) \(\frac{35^5.9^3.8^5}{81^4.32^5}=\frac{35^5.\left(3^2\right)^3.\left(2^3\right)^5}{\left(3^4\right)^4.\left(2^5\right)^5}=\frac{35^5.3^6.2^{15}}{3^{16}.2^{25}}=\frac{35^5}{3^{10}.2^{10}}=\frac{35^5}{6^{10}}\)
c) \(\frac{48^5.18^2}{81^2.34^4}=\frac{\left(2^4.3\right)^5.\left(2.3^2\right)^2}{\left(3^4\right)^2.\left(2.17\right)^4}=\frac{2^{20}.3^5.2^2.3^4}{3^8.2^4.17^4}=\frac{2^{22}.3^9}{3^8.2^4.17^4}=\frac{2^{18}.3}{17^4}\)
d) \(\frac{54^7.27^3.16^2}{243^2.64^3}=\frac{\left(2.3^3\right)^7.\left(3^3\right)^3.\left(2^4\right)^2}{\left(3^5\right)^2.\left(2^6\right)^3}=\frac{2^7.3^{21}.3^9.2^8}{3^{10}.2^{18}}=\frac{2^{15}.3^{30}}{3^{10}.2^{18}}=\frac{3^{20}}{2^3}\)
a) \(\Rightarrow3^n=3^5:3^2=3^3\)
\(\Rightarrow n=3\)
b) \(1\times2^n=4\)
\(\Rightarrow2^n=4:1=4=2^2\)
\(\Rightarrow n=2\)
c) \(3^2\times3^n=3^7\)
\(\Rightarrow3^n=3^7:3^2=3^5\)
\(\Rightarrow n=5\)
d) \(\frac{1}{9}=\left(\frac{1}{3}\right)^2\)
\(\left(\frac{1}{3}\right)^2\times27^n=3^n\)
Làm tới đây rồi khúc sau thực sự không chắc lắm
\(3^{x+2}-3^x-9^5=27^3\cdot5\)
\(\Rightarrow3^{x+2}-3^x-\left(3^2\right)^5=\left(3^3\right)^3\cdot5\)
\(\Rightarrow3^{x+2}-3^x-3^{10}=3^9\cdot5\)
\(\Rightarrow3^{x+2}-3^x=3^9\cdot5+3^{10}\)
\(\Rightarrow3^x\cdot3^2-3^x\cdot1=3^9\cdot\left(5+3\right)\)
\(\Rightarrow3^x\left(9-1\right)=3^9\cdot8\)
\(\Rightarrow3^x\cdot8=3^9\cdot8\)
\(\Rightarrow3^x=3^9\)
\(\Rightarrow x=9\)
Vậy: ...