Question

Zadanie 1
Rozwiąż korzystając ze wzorów:
(¼a+3b)²=
(⅓x²+4y)²=
(6a²b²-a³)²=
(2½a³-ab)²=

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Wzory:
(a+b)² = a² + 2ab + b²
(a-b)² = a² - 2ab + b²
a²-b² = (a - b) * (a + b)

Answer 1

(¼a+3b)²=1/16a^2+3/2ab+9b^2
(⅓x²+4y)²=1/9x^4+8/3x^2*y+16y^2
(6a²b²-a³)²=36a^4*b^4-12a^5b^2+a^6
(2½a³-ab)²=25/4a^6-5a^4*b+a^2b^2

Answer 2

Zadanie 1
Rozwiąż korzystając ze wzorów:
(¼a+3b)²= (1/4a)² + 2*1/4a*3b + (3b)² = (1/16)a² + (3/2)ab + 9b²

(⅓x²+4y)²=(1/3x²)² + 2*(1/3)x*4y + (4y)² = (1/9)x⁴+ (8/3)xy +16y²

(6a²b²-a³)²=(6a²b²)² - 2*6a²b²*a³ - (a³)² = 36a⁴b⁴ -12a⁵b² - a⁶

(2½a³-ab)²= (5/2a³)² - 2*(5/2)a³*ab + (ab)² = (25/4)a⁶ - 5a⁴b + a²b²