Question

$Demonstrati~ca ~urmatoarele~propozitii~sunt~adevarate:$[tex]a)~4+2 \sqrt{3} =( \sqrt{3} +1)^2\\
b)~7-4 \sqrt{3} =(2- \sqrt{3} )^2\\
c)~5+2 \sqrt{6} =( \sqrt{3}+ \sqrt{2})^2\\
d)~7+ \sqrt{24} =(1+ \sqrt{6})^2\\
e)~13+2 \sqrt{42}=( \sqrt{6} + \sqrt{7} )^2\\
f)~11-6 \sqrt{2} =(3- \sqrt{2} )^2\\
g)~2 \sqrt{7}+8=(\sqrt{7}+1)^2\\
h)~ \sqrt{4+ 2\sqrt{3} } = \sqrt{3} +1\\
i)~ \sqrt{7-4 \sqrt{3}}=2- \sqrt{3} \\
j) \sqrt{5+2 \sqrt{6} } = \sqrt{3}+ \sqrt{2} \\
k)~ \sqrt{7+ \sqrt{24} } =1+ \sqrt{6} \\
l)~ \sqrt{13+2 \sqrt{42} } = \sqrt{6} + \sqrt{7} \\
[/tex] 
[tex]m) ~\sqrt{11+6 \sqrt{2} } =| \sqrt{2}-3| \\
n)~ \sqrt{7+2 \sqrt{10} } = \sqrt{5}+ \sqrt{2}.\\
Vreau~sa~imi~si~explicati~!!![/tex]

Answer 1

am atasat raspunsul la primele este formula cu (a+b)²=a²+2ab+b² si al urmatoarele este ridicarea la patrat a ambelor ecuatiisi verifica toate egalitatea

Answer 2

Sper ca prin rezolvarea detaliata am reusit sa ma fac inteles ! 
Am aplicat formulele de calcul prescurtat: (a+b)² =a²+2ab+b²  si  (a-b)² =a² -2ab+b²