x^3 x^b (x^2)^b+ 1=x^29

x^(3+b+2b+2)=x^29

x^(3b+5)=x^29

∴3b+5=29

3b=24

b=8

Fractional equation (7/x2+x)-(3/x-x2) =1+(7-x2/x2-1)

(7/x^2+x)-(3/x-x^2)= 1+(7-x^2/x^2- 1)

7/x(x+ 1)+3/x(x- 1)= 1+(7-x？ )/(x- 1)(x+ 1)

[7(x- 1)+3(x+ 1)]/x(x- 1)(x+ 1)=[x(x- 1)(x+ 1)+x(7-x？ )]/x(x- 1)(x+ 1)

7x-7+3x+3=x？ -x+7x-x？

10x-6x=4

4x=4

x= 1

∫x = 1 equation is meaningless.

∴ The original equation has no solution.

Given the square of x +3x- 1 =0, find the square of the algebraic expression x-3÷3x(x-2)÷x -9÷x-2.

X-3÷3x(x-2)÷(x squared -9)÷x-2

=(x-3)/3x(x-2)÷(x-3)(x+3)÷(x-2)

= 1/3x(x-2)(x+3)÷(x-2)

= 1/3x(x+3)

= 1/(3x？ +9x-3+3)

= 1/[3(x？ +3x- 1)+3]

= 1/(0+3)

= 1/3

Let the smaller roots of the square of the equation (1996x)-1997x-1= 0 and x2+ 1995x- 1996=0 be m and n respectively.

The square of (1996x)-1995x1997x-1= 0.

( 1996x)？ -( 1996- 1)( 1996+ 1)x- 1 = 0

( 1996x)？ -( 1996? - 1)x- 1=0

( 1996x)？ - 1996? x+x- 1=0

1996? x(x- 1)+x- 1=0

(x- 1)( 1996？ x- 1)=0

∴X 1= 1

X2= 1/ 1996？ =M

x？ + 1996x- 1996=0

(x- 1)(x+ 1996)=0

∴X3= 1

X4=- 1996=N

∴M= 1/ 1996？

N=- 1996