
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


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) \(\frac{54^7.27^3.16^2}{243^2.64^3}=\frac{\left(2.3^3\right)^7.\left(3^3\right)^3.\left(2^4\right)^2}{\left(3^5\right)^2.\left(2^6\right)^3}=\frac{2^7.3^{21}.3^9.2^8}{3^{10}.2^{18}}=\frac{2^{15}.3^{30}}{3^{10}.2^{18}}=\frac{3^{20}}{2^3}\)

Ta có : 3x + 4 + 3x = 720
=> 3x(34 + 1) = 720
=> 3x.82 = 720
=> 3x = 720 : 82
=> 3x = ?
Ta có : 5x + 3 - 5x + 2 = 12500
=> 5x + 2(5 - 1) = 12500
=> 5x + 2 .4 = 12500
=> 5x + 2 = 3125
=> 5x + 2 = 55
=> x + 2 = 5
=> x = 3
Vậy x = 3

2|x - 5| = 8
\(\Rightarrow\) |x - 5| = 4
\(\Rightarrow\) |x - 5| = \(\left\{{}\begin{matrix}4\\-4\end{matrix}\right.\)
\(\Rightarrow\) x = \(\left\{{}\begin{matrix}9\\1\end{matrix}\right.\)

a. 84.165 = (23)4. (24)5 = 212. 220= 232
b. 540.1252.6253= 540. (53)2.(54)3= 540.56.512= 558
c. 274.8110= (33)4.(34)10= 312.340= 352
d. 103 . 1005 .10004= 103. (102)5.(103)4= 103.1010.1012= 1025

Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

\(x^{2018}-x^{18}=0\)
\(x^{18}.\left(x^{2018}-1\right)=0\)
\(=>\orbr{\begin{cases}x^{18}=0\\x^{2018}-1=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
b) 275 > 81x
<=> 315 > 34x
<=> 15 > 4x
<=> x < 15 /4
c) 1252+x > 258
<=> 53(2+x) > 516
<=> 3(2+x) > 16
<=> 6 + 3x > 16
<=> 3x > 10
<=> x > 10/3
d) 5x . 5x+1 . 5x+2 <= 100...0 ( 18 số 0 ) : 218
<=> 5x+x+1+x+2 <= 1018 : 218
<=> 53x+3 <= 518
<=> 3x+3 <= 18
<=> 3x <= 15
<=> x <= 5
( <= là bé hơn hoặc bằng )

Ta có \(27^5:81^5=\left(3^3\right)^5:\left(3^4\right)^5\)
\(=3^{10}:3^{20}\)
\(=3^{15-20}\)
\(=\frac{1}{3^5}\)
Ta có \(x^{10}=1^x\)
\(\Rightarrow x^{10}=1\)
\(\Rightarrow x=1\)
a)\(27^5:81^5\)
\(=\left(\frac{27}{81}\right)^5\)
\(=\left(\frac{1}{3}\right)^5=\frac{1}{3^5}\)
b)\(x^{10}=1^x\)
\(\Rightarrow x^{10}=1\)
\(\Rightarrow x=\orbr{\begin{cases}1\\-1\end{cases}}\)