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)=245482,813

b)=9.779744327x10¹⁴

tự giải nốt 

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 \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)

bạn tính ra rồi rút gọn nha

Bài 1:

Tao có:

\(81^7mod\left(405\right)\)

\(81^3\equiv81mod\left(405\right)\)

\(81^6\equiv81^2\equiv81mod\left(405\right)\)

\(81^7\equiv81^2.81\equiv81mod\left(405\right)\)

Ta có:

\(27^9mod\left(405\right)\)

\(27^3\equiv243mod\left(405\right)\)

\(27^9\equiv243^3\equiv162mod\left(405\right)\)

Ta có:

\(9^{13}mod\left(405\right)\)

\(9\equiv9mod\left(405\right)\)

\(9^3\equiv324mod\left(405\right)\)

\(9^9\equiv324^3\equiv324mod\left(405\right)\)

\(9^{10}\equiv324.9\equiv81mod\left(405\right)\)

\(9^{13}\equiv81.324\equiv324mod\left(405\right)\)

\(81^7+27^9-9^{13}:405=81+162-324:405=-0,2\)

\(\Rightarrow81^7+27^9-9^{13}⋮405\left(đpcm\right)\)

Casio không biết có áp dụng ntn vào bài này được không nữa? Nhưng mình ôn hổm rày thấy có bài gần giống vậy, nên mình làm thử bạn tham khảo nha chúc bạn học tốt! ^^

Bài 1 :

a/   \(a^3.a^9=a^{3+9}=a^{12}\)

b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)

c/  \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)

d/  \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)

e/  \(5^6:5^3+3^3.3^2\)

     \(=5^3+3^5=125+243=368\)

i/ \(4.5^2-2.3^2\)

   \(=2^2.5^2-2.3^2\)

   \(=2^2.25-2^2.14\)

   \(=2^2.\left(25-14\right)\)

   \(=2^2.11\)      

   \(=4.11=44\)

Bài 2 :

a, \(3^8:3^6=3^{8-6}=3^2\)

 \(7^5:7^2=7^{5-2}=7^3\)

 \(19^7:19^3=19^{7-3}=19^4\)

  \(2^{10}.8^3=2^{10}.\left(2^3\right)^3=2^{10}.2^9=2^{19}\)  

\(12^7:6^7=\left(12:6\right)^7=2^7=128\)

\(27^5:81^3=\left(3^3\right)^5:\left(3^4\right)^3=3^{15}:3^{12}=3^3=9\)

2^x = 32

=> 2^x = 2⁵

=> x = 5

vậy_

a) \(2^x=32\)

Ta có: \(2^5=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

b) Sửa đề tí: \(9< 3^x< 81\)

\(\Rightarrow3^2< 3^x< 3^4\)

\(\Rightarrow2< x< 4\)

\(\Rightarrow x=\left\{3\right\}\)

Vậy x = 3

c) Ta có: \(25\le5^x\le125\)

\(\Rightarrow5^2\le5^x\le5^3\)

\(\Rightarrow2\le x\le3\)

\(\Rightarrow x=\left\{2;3\right\}\)

Vậy x = 2 hoặc x = 3

d) \(\left(x-2\right)^3\times5=40\)

\(\Rightarrow\left(x-2\right)^3=8\)

Mà \(8=2^3\Rightarrow\left(x-2\right)^3=2^3\)

Suy ra: x - 2 = 2

Vậy x = 4

c/ 2x - 1 = \(5^{98}:5^{96}\)

2x - 1 = \(5^2\) = 25

2x = 25 + 1 = 26

x = 26 : 2

x = 13

d/ 7x + 3 = \(3^5.2^3.9\)

7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)

7x + 3 = \(17496\)

7x = 17496 - 3 = 17493

x = 17493 : 7

x = 2499

e/\(2^{2x+6}=1\)

\(2^{2x+6}=2^0\)

2x + 6 = 0

2x = 0 - 6 = - 6

x = - 6 : 2

x = - 3

j/ \(2^x=8\)

\(2^x=2^3\)

x = 3

g/ \(2^x:2^3=16\)

\(2^{x-3}=2^4\)

x - 3 = 4

x = 4 + 3

x = 7

h/ \(2^x+2^{x+1}+2^{x+2}=56\)

\(2^x\left(1+2+2^2\right)\) = 56

\(2^x.7=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

x = 3

Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!

a, \(x+2^3=5^2\Rightarrow x+8=25\Rightarrow x=17\)

b, \(2x+3=7^2\Rightarrow2x+3=49\Rightarrow x=\dfrac{49-3}{2}=23\)

Còn lại tự làm.

a.\(\hept{\begin{cases}4^8.2^{20}=2^{16}.2^{20}=2^{36}\\9^{12}.27^5.81^4=3^{24}.3^{15}.3^{12}=3^{51}\\64^3.4^5.16^2=2^{18}.2^{10}.2^8=2^{36}\end{cases}}\)

b.\(\hept{\begin{cases}25^{20}.125^4=5^{40}.5^{12}=5^{52}\\x^7x^4x^3=x^{14}\\3^6.4^6=12^6\end{cases}}\)

c.\(\hept{\begin{cases}8^4.2^3.16^2=2^{12}.2^3.2^8=2^{23}\\2^3.2^2.8^3=2^3.2^2.2^9=2^{14}\end{cases}}\)