\(9^7.81:9^5=9^7.9^2:9^5=9^{7+2-5}=9^4\\ x^{12}:x.x^8=x^{12-1+8}=x^{19}\\ 16.2^4:8=2^4.2^4:2^3=2^{4+4-3}=2^5\\ 64.4^5:16=4^3.4^5:4^2=4^{3+5-2}=4^6\\ 3^{12}.3:3^8=3^{12+1-8}=3^5\\ 7^9.7^{12}:2015^0=7^{9+12}:1=7^{19}\)

\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)

\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)

\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)

Vậy x = 5/19

\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

Vậy x = 1/2 hoặc x = -6

\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Vậy x = 7 hoặc x = -1

7 x 8 = 56       16 : 2 = 8       36 : 6 = 6       49 : 7 = 7

2 x 5 = 10       72 : 8 = 9       9 x 3 = 27       63 : 7 = 9

6 x 4 = 24       25 : 5 = 5       4 x 8 = 32       7 x 5 = 35

7 x 8 = 56

2 x 5 = 10

6 x 4 = 24

16 : 2 = 8 

72 : 8 = 9

25 : 5 = 5

36 : 6 = 6

9 x 3 = 27

4 x 8 = 32

49 : 7 = 7

63 : 7 = 9

7 x 5 = 35

\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)

\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)

\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)

\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
 

6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)

7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)

8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)

9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)

10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)

6) \(...=\dfrac{3}{2}.\left(\dfrac{3}{2}\right)^2.\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

7) \(...=2^{-7}.2^3.2^5.2^8=2^9\)

8) \(...=\left(\dfrac{1}{7}\right)^4.5^3.5=\left(\dfrac{1}{7}\right)^4.5^4\)

9) \(...=2^2.2^5:\left(2^3.2^{-4}\right)=2^8\)

10) \(...=\left(\dfrac{1}{7}\right)^3.\left(\dfrac{1}{7}\right)^2=\left(\dfrac{1}{7}\right)^5\)

Ta có: \(\dfrac{3x+5}{16}+\dfrac{2x+7}{8}=\dfrac{4+3x}{32}+2x\)

\(\Leftrightarrow\dfrac{2\left(3x+5\right)}{32}+\dfrac{4\left(2x+7\right)}{32}=\dfrac{4+3x}{32}+\dfrac{64x}{32}\)

Suy ra: \(6x+10+8x+28=4+3x+64x\)

\(\Leftrightarrow67x+4-14x-38=0\)

\(\Leftrightarrow53x=34\)

hay \(x=\dfrac{34}{53}\)

a: =7/8:(2/9-18+1/36)-5/12

=-7/142-5/12=-397/852

b: =3/7(4/9+5/9:6/12)=2/3

c: =5^8(16/31-47/31)+1/3=-5^8+1/3

d: =7/2(3/8+5/8:4/15)=609/64

Lời giải:

Gọi biểu thức là A.

\(A=256.\frac{1}{8}+\frac{1}{49^2}.7^3+\frac{1}{36^2}.\frac{1}{8^2}.27\\ =32+\frac{1}{7}+\frac{1}{3072}=32\frac{3079}{21504}\)

a: =>x/16=3/8+7/2=3/8+28/8=31/8=62/16

=>x=62

b: =>3/5:x=17/10-2/5=17/10-4/10=13/10

=>x=3/5:13/10=3/5x10/13=30/65=6/13