Question

Zadanie 2,3 , 2,4,2,7 jak można :)

Answer 1

W zalaczniku rozwiazanie

Answer 2

2,3
(x - 4)² - (x - 3)(x + 3) + 2(x² + 5) + 5x =
= x² - 8x + 16 - x² + 9 + 2x² + 10 + 5x =
= 2x² - 3x + 35
Podstawiam:
2* (-1)² - 3 * (-1) + 35 =
= 2 + 3 + 35 =
= 40

2,4
2(5x² + 4) + (2x - 3) (2x + 3) -2(3x² - 1) + 3(1 - 2x)² =
= 10x² + 8 + 4x² - 9 - 6x² + 2 +3(1 - 4x + 4x²) =
= 8x² + 1 + 3 - 12x + 12x² =
= 20x² - 12x + 4
Podstawiam:
20*(1/2)² - 12* 1/2 + 4 =
= 5 - 6 + 4 =
= 3

2,7
4[(1/2x + y)² - (1/2x - y)²] - y(2/3x + 5) =
= 4* (1/4x² + xy + y² - 1/4x² - xy + y²) - 2/3xy -5y =
= 4* 2y² - 2/3xy - 5y =
= 8y² - 2/3xy - 5y
Podstawiam:
8* 3,5² - 2/3* 3,5 * (-0,6) - 5* (-0,6) =
= 8* 12,25 + 1,4 + 3 =
= 98 + 4,4 =
= 102,4