Question

HELP PLEASE !!! Place each expression under the equivalent expression in the table.

Answer 1

See the attached figure.
Note: the remaining items can not be put in the table

The detailed solution is as shown in the figures.

Answer 2

Option 3,5 are under first expression and Option 1,2 are under second expression.

Explanation:

1).
`((x-4)(x+2))/(x^(2)+5x+6)+(-3x^(2)+24x-20)/((x+3)(4x-5))`

`=((x-4)(x+2))/((x+3)(x+2))+(-3x^(2)+24x-20)/((x+3)(4x-5))`
`=((x-4))/((x+3))+(-3x^(2)+24x-20)/((x+3)(4x-5))`

`=((x-4)(4x-5)-3x^(2)+24x-20)/((x+3)(4x-5))`

`=(4x^(2)-5x-16x+20-3x^(2)+24x-20)/((x+3)(4x-5))`
`=(x^(2)+3x)/((x+3)(4x-5))`

`=(x(x+3))/((x+3)(4x-5))=(x)/(4x-5)`

2).
`(3x)/(4x-5)-(4x^(2))/(8x^(2)-10x)=(3x)/(4x-5)-(4x)/(8x-10)`
`=(3x(8x-10)-4x(4x-5))/((4x-5)(8x-10))`

`=(24x^(2)-30x-16x^(2)+20x)/((4x-5)(8x-10))`

`=(8x^(2)-10x)/((4x-5)(8x-10))`
`=(x(8x-10))/((4x-5)(8x-10))=(x)/(4x-5)`

3).
`(6x^(2))/(x^(2)-7x+10)/ (2x)/(x-5)`

`=(6x^(2))/(x^(2)-7x+10)* (x-5)/(2x)`

`=(3x(x-5))/(x^(2)-7x+10)`

`=(3x(x-5))/((x-5)(x-2))=(3x)/(x-2)`

4).
`(-x)/(4x-5)-(4x^(2))/(16x^(2)-22x)`

`=(-x)/(4x-5)-(2x)/(8x-11)`

`=(-x(8x-11)-2x(4x-5))/((4x-5)(8x-11))`

`=(-8x^(2)+11x-8x^(2)+10x)/((4x-5)(8x-11))`

`=(-16x^(2)+21x)/((4x-5)(8x-11))=(-x(16x-21))/((4x-5)(8x-11))`

5).
`(3x^(2))/(x+3)* (2(x+3))/(2x^(2)-4x)`

`=(6x^(2)(x+3))/(2x(x+3)(x-2))=(3x)/(x-2)`

6).
`(5x^(2))/(x-2)* (2x+6)/(8x^(2)-4x)`

`=(10x^(2)(x+3))/(4x(x-2)(2x-1))`