Question

alguém poderia me ajudar? obrigado.

Answer 1

a) $\sqrt{3} + \sqrt{27} =$

27 / 3
9 / 3
3 / 3
1          --------- $\sqrt{}$3$^{2} . 3$ ---- 3$\sqrt{3 }$

$1 \sqrt{3} + 3 \sqrt{3} = 4 \sqrt{3}$

b) $\sqrt{75} = [tex] \sqrt{3 ^{2}.5 } = 3 \sqrt{5}$
    $\sqrt{12} = [tex] \sqrt{2 ^{2}.3 } = 2 \sqrt{3}$
$5 \sqrt{3} + 2 \sqrt{3} = 7 \sqrt{3}$

c) $\sqrt{50} = \sqrt{2.5.5} = 5 \sqrt{2}$
5$\sqrt{2} + 7 \sqrt{2} = 12 \sqrt{2}$

d) $\sqrt{12} = 2 \sqrt{3}$
$\sqrt{75} = 5 \sqrt{3}$
$2 \sqrt{3 } - 5 \sqrt{3} + \sqrt{3} = - 2 \sqrt{3}$

e) $3 \sqrt{20 } =(3) . 2 \sqrt{5} = 6 \sqrt{5}$
$\sqrt{32} = \sqrt{2 ^{2}. 2 ^{2}.2 } = 4 \sqrt{2}$
$- 2 \sqrt{45} = \sqrt{3 ^{2}. 5 } = (- 2). 3 \sqrt{5} = - 6 \sqrt{5}$
$\sqrt{50} = \sqrt{5 ^{2}.2 } = 5 \sqrt{2}$

 6$\sqrt{5}$ + 4$\sqrt{2}$ $- 6 \sqrt{5}$ + 2$\sqrt{5} =$
$4 \sqrt{2} + 2 \sqrt{5}$

f) $\sqrt{125} = \sqrt{5 ^{2}.2 } = 5 \sqrt{2}$
$\sqrt{27} = \sqrt{3 ^{2}.3 } =(2). 3 \sqrt{3}$ = $6 \sqrt{3}$
$\sqrt{20} = \sqrt{2 ^{2}. 5 } = 2 \sqrt{5}$
$3 \sqrt{12} = (3). \sqrt{2 ^{2}.3 } = (3).2 \sqrt{3} = 6 \sqrt{3}$

somando e diminuindo fica : $5 \sqrt{2} + 12 \sqrt{3} - 2 \sqrt{5}$