Question

(3-√2)^2
(2√2+3)^2
(4-√3)•(4+√3)
Pascu pas va rog​

Answer 1

Vom folosi formulele de calcul prescurtat :

(a-b)^2=a^2+b^2-2ab

(a+b)^2=a^2+b^2+2ab

(a-b)(a+b)=a^2-b^2

1. $(3-\sqrt{2})^2 = 3^2+\sqrt{2}^2-2*3*\sqrt{2}=9+2-6*\sqrt{2}=11-\sqrt{36}*\sqrt{2}=11-\sqrt{72}$

2. $(2\sqrt{2}+3)^2 =(\sqrt{2^2*2}+3)^2 = (\sqrt{8}+3)^2=\sqrt{8}^2+3^2+2*3*\sqrt{8}=8+6\sqrt{8}=8 + \sqrt{6^2*8}=8+\sqrt{288}$

3.$(4-\sqrt{3})(4+\sqrt{3})=4^2-\sqrt{3}^2=16-3=13$

Answer 2

(3 - V2)^2 = 3^2 - 2x3x(-V2) + (-V2)^2 = 9 + 6V2 + 2 = 11 + 6V2

(2V2 + 3)^2 = (2V2)^2 + 2x2V2x3 + 3^2 = 4x2 + 12V2 + 9 = 17  + 12V2

(4-V3)(4+V3) = 4^2 - (V3)^2 = 16 - 3 = 13

Sfat: invata regulile de calcul prescurtat!