Question

Vă rog problemele 1 si 2.... urgenttt

Answer 1

1) a)
a = 4 + 3√2 - 4, |3√2 - 4| = 3√2 - 4 ,deoarece 3√2 > 4 (18 > 16); => a = 3√2;
 b = 7√2 - 3√2 => b = 4√2;
presupunem : a < b <=> 3√2 < 4√2 | : √2 => 3 < 4 ,,A" => a < b;
b) ma = (a + b)/2 => ma = (3√2 + 4√2)/2 => ma = 7√2 / 2;
c) √a*b = √3√2*4√2 = √12 * 2 = √24 = 2√6;
2. Aducem la numitor comun in paranteza si obtinem (numitor comun 48) :
(4√12 - 16 * 2√3 + 3√48)/48 *√108/5  + 1 = 
=(4 * 2√3 - 32√3 + 3*4√3)/48 * 6√3/5 + 1
= (20√3 - 32√3)/48 * 6√3/5 + 1
=(-12√3)/48 * 6√3/5 + 1
=-√3/4* 6√3/5 + 1
= -6*(√3)² / 20 + 1
= (-6) * 3 / 20 + 1 
= -18 / 20 + 1
= -9 / 10 + 1
= (-9 + 10) / 10
=1/10;

Answer 2

poza!
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