x/y=3

x=3y

(x^2+xy)/y^2

=[(3y)^2+3y*y]/y^2

=(9+3)y^2/y^2

= 12

24B 2-9A2 divided by 2B 39AB-3B2.

[(b^2-9a^2)/(4b^2)]/[(9ab-3b^2)/(2b^3)]

=[(b+3a)(b-3a)/(4b^2)×(2b^3)/[3b(3a-b)]

=-3b(b+3a)*b/2

=-3/2b^3-9/2ab^2

3 x 2+4x+4 x 2-6x+9 divided by x 2-9 multiplied by x 2+3x of 3-x, where x=2, is decomposed first and then evaluated.

[(x^2-6x+9)/(x^2+4x+4)]/[(x^2-9)/x]×[(x^2+3x)/(3-x)]

=[(x-3)^2/(x+2)^2]×x/[(x+3)(x-3)]×[-x(x+3)/(x-3)]

=-x^2/(x+2)^2

=-[x/(x+2)]^2

=-[2/(2+2)]^2

=- 1/4

4 (x 2-y 2) in 4(xy) 2 is divided by (x+y) 2 and then multiplied by (x) in x-y x) 3, and then decomposed and evaluated.

[(x^2-y^2)/(xy)]^2/(x+y)^2×[x/(x-y)]^3

=[(x+y)^2(x-y)^2/(xy)^2]/(x+y)^2×[x^3/(x-y)^3]

=x/[y^2(x-y)]

=3/[2^2(3-2)]

=3/4

5 Given that x 2' s 3x+4-x-2 = x-2' s A-x+ 1' s B and AB are constants, find the value of 4A-B..

(3x+4)/(x^2-x-2)=a/(x-2)-b/(x+ 1)

(3x+4)/(x^2-x-2)=[a(x+ 1)-b(x-2)]/(x^2-x-2)

(3x+4)/(x^2-x-2)=[(a-b)x+(a+2b)]/(x^2-x-2)

A-B=3..........( 1)

A+2B=4.......(2)

(2)-( 1):3B= 1

B= 1/3

Substitution (1): A- 1/3 = 3.

A= 10/3

4A-B

=4× 10/3- 1/3

=(40- 1)/3

= 13