Question

1) simplifique os radicais e efetue as operações:

a) √2+√32=
b) √27+√3=
c) 3√5+√20=
d) √27+5√3=
e) 2√7+√28=
f) √50-√98=
g) √20-√45
h) √28-10√7=
i) 9√2+3√50=
j) 6√3+√75=
k) √98+5√18=
l) 3√98-2√50=
m) 3√8-7√50=
n) 2√32-5√18=
o) √75-2√12+√27=
p) √12-9√3+√75=
q) √98-√18-5√32=
r) 5√180+√245-17√5=

alguém me ajudaa pfvr

Answer 1

Explicação passo-a-passo:

a) $\sqrt{2}+ \sqrt{32}=\sqrt{2}+ 4\sqrt{2}=5\sqrt{2$

b)$\sqrt{27}+\sqrt{3}=3\sqrt{3}+\sqrt{3}=4\sqrt{3}$

c) $3\sqrt{5}+\sqrt{20}=3\sqrt{5}+2\sqrt{5}=5\sqrt{5}$

d) $\sqrt{27}+5\sqrt{3}=3\sqrt{3}+5\sqrt{3}=8\sqrt{3}$

e) $2\sqrt{7}+\sqrt{28}=2\sqrt{7}+2\sqrt{7}=4\sqrt{7}$

f) $\sqrt{50}-\sqrt{98}=5\sqrt{2}-7\sqrt{2}=-2\sqrt{2}$

g) $\sqrt{20}-\sqrt{45}=2\sqrt{5}-3\sqrt{5}=-\sqrt{5}$

h) $\sqrt{28}-10\sqrt{7}=2\sqrt{7}-10\sqrt{7}=-8\sqrt{7}$

i) $9\sqrt{2}+3\sqrt{50}=9\sqrt{2}+3\sqrt{50}=9\sqrt{2}+3*(5\sqrt{2})=9\sqrt{2}+15\sqrt{2}=24\sqrt{2}$

j) $6\sqrt{3}+\sqrt{75}=6\sqrt{3}+5\sqrt{3}=11\sqrt{3}$

k) $\sqrt{98}+5\sqrt{18}=\sqrt{98}+5\sqrt{18}=7\sqrt{2}+5(3\sqrt{2})=7\sqrt{2}+15\sqrt{2}=22\sqrt{2}$

l) $3\sqrt{98}-2\sqrt{50}=3(7\sqrt{2})-2(5\sqrt{2})=21\sqrt{2}-10\sqrt{2}=11\sqrt{2}$

m) $3\sqrt{8}-7\sqrt{50}=3(2\sqrt{2})-7(5\sqrt{2})=6\sqrt{8}-35\sqrt{2}=-29\sqrt{2}$

n) $2\sqrt{32}-5\sqrt{18}=2(4\sqrt{2})-5(3\sqrt{2})=8\sqrt{2}-15\sqrt{2}=-7\sqrt{2}$

o) $\sqrt{75}-2\sqrt{12}+\sqrt{27}=5\sqrt{3}-2(2\sqrt{3})+3\sqrt{3}=5\sqrt{3}-4\sqrt{3}+3\sqrt{3}=4\sqrt{3}$

p) $\sqrt{12}-9\sqrt{3}+\sqrt{75}=2\sqrt{3}-9\sqrt{3}+5\sqrt{3}=-2\sqrt{3}$

q) $\sqrt{98}-\sqrt{18}-5\sqrt{32}=7\sqrt{2}-3\sqrt{2}+5(4\sqrt{2})=7\sqrt{2}-3\sqrt{2}+20\sqrt{2}=24\sqrt{2}$

r) $5\sqrt{180}+\sqrt{245}-17\sqrt{5}=6\sqrt{5}+7\sqrt{5}-17\sqrt{5}=-4\sqrt{5}$