47. Найдите значение выражения: 1) (16x² − 4y²)/(6x − 3y) при x = 2,5, y = −2; 2) (49c² − 9)/(49c² + 42c + 9) при c = −4.

1) \(\dfrac{16x^2-4y^2}{6x-3y}=\dfrac{4(2x-y)(2x+y)}{3(2x-y)}=\dfrac{4(2x+y)}3\). При \(x=2.5\), \(y=-2\): \(\dfrac{4(5-2)}3=4\).

2) \(\dfrac{49c^2-9}{49c^2+42c+9}=\dfrac{(7c-3)(7c+3)}{(7c+3)^2}=\dfrac{7c-3}{7c+3}\). При \(c=-4\): \(\dfrac{-31}{-25}=\dfrac{31}{25}\).

Ответ: 1) \(4\); 2) \(\dfrac{31}{25}\).