Question

calculati mg a numerelor a= √3 + 1 supara 2+√3 si b= √64 -2√2+√8
Pls ajutoorrrr

Answer 1

a=√3+(2-√3)/(4-3) = √3+2-√3 = 2 ⇒ a=2
b=8-2√2+2√2=8⇒b=8

Mg= √a*√b= √2*√8=√16=4

Answer 2

a = √3+[1/(2+√3)]

[tex]\it a=\sqrt3+\dfrac{1}{2+\sqrt3} =\sqrt3+\dfrac{2-\sqrt3}{2^2-(\sqrt3)^2}=\sqrt3+\dfrac{2-\sqrt3}{4-3}= \\\;\\ =\sqrt3+2-\sqrt3=2[/tex]

$\it b= \sqrt{64} -2\sqrt2+\sqrt8=8-2\sqrt2 =\sqrt{4\cdot2} =8-2\sqrt2+2\sqrt2=8$

$\it m_g =\sqrt{a\cdot b} =\sqrt{2\cdot8 }=\sqrt{16}=4$