Відповідь:

[tex]\displaystyle 1)\;\frac{16x^2-4y^2}{6x-3y}=\frac{4(2x-y)(2x+y)}{3(2x-y)}=\frac{4(2x+y)}{3}=\Big[x=2.5,\; y=-2\Big]=\\\\\\=\frac{4(2\cdot2.5+(-2))}{3}=\frac{4(5-2)}{3}=\frac{12}{3}=4[/tex]

[tex]2)\;\displaystyle \frac{49c^2-9}{49c^2+42c+9}=\frac{(7c-3)(7c+3)}{(7c+3)^2}=\frac{7c-3}{7c+3}=\Big[c=-4\Big]=\frac{7\cdot(-4)-3}{7\cdot(-4)+3}=\\\\\\=\frac{-31}{-25}=\frac{31}{25}=1\frac{6}{25}[/tex]

Объяснение:

1)

(16х² - 4у²)/(6х - 3у) = ( 4(4х² - у² )/( 3(2х - у ) =

( 4(2х - у)(2х + у) )/( 3(2х - у) ) =

( 4(2х + у) )/3 = (8х + 4у)/3

х = 2,5; у = -2

(8х + 4у)/3 = (8*2,5 + 4*(-2))/3 = (20 - 8)/3 =

12/3 = 4

2)

(49с² - 9)/(49с² + 42с + 9) =

( (7с - 3)(7с + 3) )/( 7с + 3)² ) = (7с - 3)/(7с + 3)

с = -4

(7с - 3)/(7с + 3) = ( 7*(-4) - 3 )/( 7*(-4) + 3) =

(-28 - 3)/(-28 + 3) = (-31)/(-25) = 31/25 = 1 6/25 или 1,24