Question

VA ROG SA REZOLVAȚI AMBELE SUBIECTE ESTE URGENT DAU COROANA!!!​

Answer 1

Formule:

(a + b)² = a² + 2ab + b²

(a + b)(a - b) = a² - b²

SUBIECTUL 1

Ex.1

(3x + 2)² - 12x = 9x² + 12x + 4 - 12x = 9x² + 4

Ex.2

(2x - 5)(2x + 5) - 4x(x - 6)= 4x² - 25 - 4x² - 24x = -25 -24x

Ex.3

(x - 3)² + 3(2x - 3) = x² - 6x + 9 + 6x - 9 = x²

Ex.4

4(x + 1)² - (2x + 3)(2x - 3) = 4(x² + 2x + 1) - 4x² - 9 = 4x² + 8x + 4 - 4x² - 9 = 8x - 5

Ex.5

(3x + 2)(3x - 2) - 9(x + 1)(x - 1) = 9x² - 4 - 9(x² - 1) = 9x² - 4 - 9x² + 9 = 5

Ex.6

($\sqrt{2}$ + $\sqrt{3}$)² - ($\sqrt{6}$ + 1)² = 2 + 2$\sqrt{6}$ + 3 - 6 + 2$\sqrt{6}$ + 1 = 4$\sqrt{6}$

SUBIECTUL 2

Ex.1

($\sqrt{3}$ + $\sqrt{5}$)² + ($\sqrt{15}$ - 1)² = 3 + 2$\sqrt{15}$ + 5 + 15 - 2$\sqrt{15}$ + 1 = 24(B)

Ex.2

(x² + x)² - x²(x - 2)² = $x^{4}$ + 2x³ + x² - x²(x² - 4x + 4) = $x^{4}$ + 2x³ + x² - $x^{4}$ + 4x³ - 4x² = 6x³ + 5x² (C)

Ex.3

Scuze nu stiu ex.3

Ex.4

[x + ($\sqrt{3}$ - 1) ][x - ($\sqrt{3}$ - 1) ] - (x + 2)(x - 2) = x² - $(\sqrt{3} -1)^{2}$ - x² + 4 = 3 + 2$\sqrt{3}$ + 1 +  4 = 8 + $2\sqrt{3}$ (A)