Question

Ma ajuatati si pe mine cu 2 exerciti simple ?? Urgentttt !!
P.S:(poza)
Ex: Radical din 243=3 la a 5-a=3 la a4-a x 3=9 radical din 3

Answer 1

7.a)[tex]9 \sqrt{3 } [/tex]

b)$\sqrt{363} = \sqrt{ 11^{2} *3} =11 \sqrt{3}$

c)$\sqrt{507}= \sqrt{3* 13^{2} } =13 \sqrt{3}$

d) $\sqrt{1352}= \sqrt{ 2^{3}* 13 ^{2} } =2*13 \sqrt{2} =26\sqrt{2}$

e)$\sqrt{47,32}= \sqrt{ \frac{4732}{100} } =\sqrt{ \frac{ 2^{2} *1183}{100} } = \frac{2}{10} \sqrt{1183} =0,2 \sqrt{1183}$

f)$\sqrt{14,45} = \sqrt{ \frac{1445}{100} } = \sqrt{ \frac{5*17 ^{2} }{100} } = \frac{17}{10} \sqrt{5}=1,7 \sqrt{5}$

g)$\sqrt{5 x^{2} } =x \sqrt{5}$,$x$∈$Q_{+}$

h)$\sqrt{3x y^{2} } =y \sqrt{3x}$,$x$∈$Q_{+}$ si $y$∈$Q_{+}$

8.a) $2 \sqrt{2}$ sau $\sqrt{18}$?
$\sqrt{18} = \sqrt{2*3 ^{2} } =3 \sqrt{2}$
$2 \sqrt{2} < 3 \sqrt{2}$

b)$5 \sqrt{243}$ sau $27 \sqrt{27}$?
$5 \sqrt{243} =5* \sqrt{3 ^{5} } =5*9 \sqrt{3} =45 \sqrt{3}$
$27 \sqrt{27}=27 \sqrt{ 3^{3} } =27*3 \sqrt{3} =81\sqrt{3}$
$5 \sqrt{243}$ < $27 \sqrt{27}$

c) $\sqrt{75}$ sau $2 \sqrt{50}$?
$\sqrt{75}= \sqrt{3*5 ^{2} } =5 \sqrt{3}$
$2 \sqrt{50}=2* \sqrt{2*5^{2} } =2*5 \sqrt{2}=10 \sqrt{2}$
$\sqrt{75}$<$2 \sqrt{50}$