\(i)4^8:x=4^6\\ x=4^8:4^6\\ x=4^2\\ k)12x-33=3^5\\ 12x-33=243\\ 12x=243+3\\ 12x=276\\ x=276:12\\ x=23\\ l)\left(5x+335\right):2=20^2\\ \left(5x+335\right):2=400\\ 5x+335=400.2\\ 5x+335=800\\ 5x=800-335\\ 5x=465\\ x=465:5\\ x=93\) 

\(m)\left(x^2-10\right):5=3\\ x^2-10=3.5\\ x^2-10=15\\ x^2=15+10\\ x^2=25\\ x^2=5^2\\ 740:\left(x+10\right)=10^2-2.13\\ 740:\left(x+10\right)=100-26\\ 740:\left(x+10\right)=74\\ x+10=740:74\\ x+10=10\\ x=10-10\\ x=0.\)

i) 4⁸:x=4⁶

<=> x = 4⁸ : 4⁶ = 4² = 16

k)12x-33=3⁵

<=> x = (3⁵ + 33) : 12 = 23

l)(5x+335):2=20²

<=> x = (20² x 2 - 335) : 5 = 93

m)(x²-10):5=3

<=> x² = 3 x 5 + 10 = 25

<=> x = 5 hoặc x = -5

740:(x+10)=10²-2.13

<=> x = 740 : (10² - 2.13) - 10 = 0

=> 6x + 3(10 - x) = 45

=> 6x + 30 - 3x = 45

=> 3x + 30 = 45

=> 3x = 15

=> x = 5

Đề bài là j vậy ak ??

Đề bài ???????

\(\frac{-45}{91}< \frac{777}{999}\)vì:

\(\frac{-45}{91}< 1< \frac{777}{999}\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=\dfrac{6}{2}n_{Al}=0,6mol\\ n_{AlCl_3}=n_{Al}=0,2mol\\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol\\ C_{\%HCl}=\dfrac{0,6.36,5}{500}\cdot100\%=4,38\%\\ C_{\%AlCl_3}=\dfrac{0,2.133,5}{5,4+500-0,3.2}\cdot100\%=5,29\%\)

x/5=-3/14.7/3

x/5=-1/2

x   =-1/2.5

x   =-5/2.Vậy x=-5/2