Odpowiedź:
Ćwiczenie 4
zastosowany wzór
(a - b)² = a² - 2ab + b²
a)
(x - 2)² = x² - 4x + 4
b)
(2x - 7)² = 4x² - 28x + 49
c)
(4 - 3x)² = 16 - 24x + 9x² = 9x² - 24x + 16
d)
(1/2y - 5x)² = 1/4y² - 10/2xy + 25x² = 25x² - 5xy + 1/4y²
e)
(√3 - 1)² = 3 - 2√3 + 1 = 2 - 2√3 = 2(1 - √3)
zad 5.4
a)
(2 - x)² = 4 - 4x + x² = x² - 4x + 4
b)
(4 - x)² = 16 - 8x + x² = x² - 8x + 16
c)
(5z - 1)² = 25z² - 10z + 1
d)
(1 - xy)² = 1 - 2xy + x²y² = x²y² - 2xy + 1
e)
(3x - 2y)² = 9 x² - 12xy + 4y²
f)
(4xy - 3)² = 16x²y² - 24xy + 9
g)
(1/4x - 3y)² = 1/16x² - 6/4xy + 9y² = 1/16x² - 1 1/2xy + 9y²
h)
(- 2 - x)² = - (2 + x)² = - (4 + 4x + x²) = - x² - 4x - 4
