Question

Ajutatima va rog dau steluța !!
a) (√12+1)²
b)(√3+√2)²
c) (√8-3)²
d)(√10-√6)²

Answer 1

a) (√12+1)²= 12+2√12+1= 13+4√3
b)(√3+√2)²=3+2√6+2= 5+2√6
c) (√8-3)²= 8-6√4*2+9= 17-12√2
d)(√10-√6)² = 10-2√60+6=16-4√15

se aplica formula (a+b)²= a²+2ab+b² si formula (a-b)²= a²-2ab+b².

Answer 2

$\displaystyle a).\left( \sqrt{12} +1\right)^2=\left( 2 \sqrt{3} +1 \right)^2=\left(2 \sqrt{3} \right)^2+2 \cdot 2 \sqrt{3} \cdot 1+1^2= \\ =12+4 \sqrt{3} +1=13+4 \sqrt{3}$
$\displaystyle b).\left( \sqrt{3} + \sqrt{2} \right)^2=\left( \sqrt{3} \right)^2+2 \cdot \sqrt{3} \cdot \sqrt{2} + \left( \sqrt{2} \right)^2= \\ =3+2 \sqrt{6} +2=5+2 \sqrt{6}$
$\displaystyle c).\left( \sqrt{8} -3\right)^2=\left(2 \sqrt{2} -3\right)^2=\left(2 \sqrt{2} \right)^2-2 \cdot 2 \sqrt{2} \cdot 3+3^2= \\ =8-12 \sqrt{2} +9=17-12 \sqrt{2}$
$\displaystyle d).\left( \sqrt{10} - \sqrt{6} \right)^2=\left( \sqrt{10} \right)^2-2 \cdot \sqrt{10} \cdot \sqrt{6} + \left( \sqrt{6} \right)^2= \\ =10-2 \sqrt{60} +6=16-2 \cdot 2 \sqrt{15} =16-4 \sqrt{15}$