\(=\left(2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9\right):\left(2^{12}.\left(3^2\right)^3.31\right)\)

\(=\left(2^{13}.3^6+2^{11}.3^9\right):\left(2^{12}.3^6.31\right)\)

\(=\left[2^{11}.3^6\left(2^2+3^3\right)\right]:\left(2^{12}.3^6.31\right)\)

\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)

Đưa về phân số:

\(=\frac{2.8^4.27^2+4.6^9}{2^{12}.9^3.31}\)

\(=\frac{2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9}{2^{12}.\left(3^2\right)^3.31}\)

\(=\frac{2.2^{3.4}.3^{3.2}+2^2.2^9.3^9}{2^{12}.3^{2.3}.31}\)

\(=\frac{2.2^{12}.3^6+2^{2+9}.3^9}{2^{12}.3^6.31}\)

\(=\frac{2^{1+12}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)

\(=\frac{2^{13}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)

\(=\frac{2^{11}.3^6\left(2^2+3^3\right)}{2^{12}.3^6.31}\)

\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)

Em hiểu hơn ko?

   a) 25 . 8 ³ -  22 . 8³   

\(=8^3.\left(25-22\right)=8^3.3=1536\)

 b) 5⁴ - 2 . 5³

^{\(=5^3\left(5-2\right)=5^3.3=375\)}

^{\(c,\frac{\left(3^4+3^3\right)\left(3^4+3\right)}{9^2}=\frac{108.84}{81}=112\)}

   d) 3³. 22 - 3³. 19

\(=3^3.\left(22-19\right)=3^3.3=3^4=81\)

 e) 2⁴ . 5 - [ 131 - ( 13- 4²) ] 

\(=80-\left(131+3\right)\)

\(=-54\)

2) Tìm x 

a) 30- 5 . ( x - 3 ) = 5

\(5\left(x-3\right)=25\)

\(x-3=5\)

\(x=8\)

b) 9 + 3 . x  = 3⁷ : 3² 

\(9+3x=243\)

\(3x=234\)

x=78

C2 :

\(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^{2019}\)

\(2\cdot3\cdot3^{x+1}+4\cdot3^{x+1}=10\cdot3^{2019}\)

\(6\cdot3^{x+1}+4\cdot3^{x+1}=10\cdot3^{2019}\)

\(\left(6+4\right)\cdot3^{x+1}=10\cdot3^{2019}\)

\(10\cdot3^{x+1}=10\cdot3^{2019}\)

\(\Rightarrow x+1=2019\)

\(x=2019-1\)

\(x=2018\)

Vậy x = 2018

Chắc sai =))

\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=1\)

\(x=\frac{1313}{1212}:1\)

\(x=\frac{13}{12}\)

Lời giải 

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)

\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)

\(x=\frac{1313}{1212}:1\)

\(x=\frac{13}{12}\)

số số hạng = (x-1):2 +1 = (x+1)/2
Vài giây trước

2) 1+ 4+ 7+ 10+...+x= 330

tổng = (x+1) . (x+1)/2 : 2 = 25

=> (X+1)^2 = 100

=> x = 9

nhầm 

1) số số hạng = (x-1):2 +1 = (x+1)/2

tổng = (x+1) . (x+1)/2 : 2 = 25

=> (X+1)^2 = 100

=> x = 9

49 - 3 . ( x + 6 ) = 13

3 . ( x + 6 ) = 49 - 13

3 . ( x + 6 ) = 36

x + 6 = 36 : 3

x + 6 = 12

      x = 12 - 6

      x = 6

49 - 3.( x + 6 ) = 13

   3.( x + 6 ) = 36

     x + 6 = 12

    x = 6

Vậy x = 6

#)Giải :

\(9^8:\left(27^3.81^2\right)=\left(3^2\right)^8:\left[\left(3^3\right)^3.\left(3^4\right)^2\right]=3^{16}:\left[3^9.3^8\right]\)

Đến đây ez rùi ^^

9⁸:(27³.81²) = 9⁸:(3^3x3.3^4x2)=9⁸:3¹⁷= \(\frac{1}{3^9}\)

\(9^8:\left(27^3.81^2\right)=3^{16}:\left(3^9.3^8\right)=3^{16}:\left(3^{9+8}\right)=3^{16}:3^{17}=3^{-1}=\frac{1}{3}\)

\(9^8:\left(27^3.81^2\right)\)

\(=3^{16}:\left(3^9.3^8\right)\)

\(=3^{16}:3^{17}\)

\(=\frac{1}{3}\)