Question

Aratati ca urmatoarele propoziti sunt adevarate:
a)6+2√5=(1+√5)²
b)32-10√7=(5-√7)²
c) √16-6√7(tot sub radical)=3-√7
d)√10-2√21(tot sub radical)=√7-√3
e)√6-2√5(tot sub radical)=I1-√5I
Va rog rezolvarile complete.

Answer 1

6+2√5=(1 +√5)² =

(1+√5)² = 1²+2√5+5 = 6+2√5

32-10√7 = (5-√7)² = 32-10√7

(5-√7)²=25-10√7-7

Answer 2

   
[tex]Folosim \;formulele: \\ a^{2}+2ab+c^{2} = (a+b)^{2} \\ a^{2}-2ab+c^{2} = (a-b)^{2} \\ \\ a)\;\;\;6+2\sqrt{5}=1+2\sqrt{5}+5=1^{2}+2*1*\sqrt{5}+ (\sqrt{5})^{2} =(1+\sqrt{5})^{2} \\ b)\;\;\;32-10\sqrt{7}=25-10\sqrt{7}+7= 5^{2}-2*5*\sqrt{7}+ (\sqrt{7})^{2}=(5-\sqrt{7})^{2} \\ c)\;\;\; \sqrt{16-6 \sqrt{7}}= \sqrt{9-6\sqrt{7}+7}= \sqrt{(3-\sqrt{7})^{2}}=3-\sqrt{7} \\ d)\;\;\; \sqrt{10-2\sqrt{21}}=\sqrt{7-2\sqrt{7}\sqrt{3}+3}=\sqrt{(\sqrt{7}-\sqrt{3})^{2}}=\sqrt{7}-\sqrt{3} [/tex]

$d)\;\;\; \sqrt{6-2 \sqrt{5}} =\sqrt{5-2 \sqrt{5}+1}=\sqrt{(\sqrt{5}-1)^{2}}=\sqrt{5}-1$