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

Answer:

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

Step-by-step explanation:

1). $\frac{(x-4)(x+2)}{x^{2}+5x+6}+\frac{-3x^{2}+24x-20}{(x+3)(4x-5)}$

$=\frac{(x-4)(x+2)}{(x+3)(x+2)}+\frac{-3x^{2}+24x-20}{(x+3)(4x-5)}$$=\frac{(x-4)}{(x+3)}+\frac{-3x^{2}+24x-20}{(x+3)(4x-5)}$

$=\frac{(x-4)(4x-5)-3x^{2}+24x-20}{(x+3)(4x-5)}$

$=\frac{4x^{2}-5x-16x+20-3x^{2}+24x-20}{(x+3)(4x-5)}$$=\frac{x^{2}+3x}{(x+3)(4x-5)}$

$=\frac{x(x+3)}{(x+3)(4x-5)}=\frac{x}{4x-5}$

2). $\frac{3x}{4x-5}-\frac{4x^{2}}{8x^{2}-10x}=\frac{3x}{4x-5}-\frac{4x}{8x-10}$$=\frac{3x(8x-10)-4x(4x-5)}{(4x-5)(8x-10)}$

$=\frac{24x^{2}-30x-16x^{2}+20x}{(4x-5)(8x-10)}$

$=\frac{8x^{2}-10x}{(4x-5)(8x-10)}$$=\frac{x(8x-10)}{(4x-5)(8x-10)}=\frac{x}{4x-5}$

3). $\frac{6x^{2}}{x^{2}-7x+10}\div \frac{2x}{x-5}$

$=\frac{6x^{2}}{x^{2}-7x+10}\times \frac{x-5}{2x}$

$=\frac{3x(x-5)}{x^{2}-7x+10}$

$=\frac{3x(x-5)}{(x-5)(x-2)}=\frac{3x}{x-2}$

4). $\frac{-x}{4x-5}-\frac{4x^{2}}{16x^{2}-22x}$

$=\frac{-x}{4x-5}-\frac{2x}{8x-11}$

$=\frac{-x(8x-11)-2x(4x-5)}{(4x-5)(8x-11)}$

$=\frac{-8x^{2}+11x-8x^{2}+10x}{(4x-5)(8x-11)}$

$=\frac{-16x^{2}+21x}{(4x-5)(8x-11)}=\frac{-x(16x-21)}{(4x-5)(8x-11)}$

5). $\frac{3x^{2}}{x+3}\times \frac{2(x+3)}{2x^{2}-4x}$

$=\frac{6x^{2}(x+3)}{2x(x+3)(x-2)}=\frac{3x}{x-2}$

6). $\frac{5x^{2}}{x-2}\times \frac{2x+6}{8x^{2}-4x}$

$=\frac{10x^{2}(x+3)}{4x(x-2)(2x-1)}$