Question

sa se arate ca numarul  rad(7+4rad3)-rad3 este natural

Answer 1

√(7+4√3) = √(7+48)

A=7
B=48
C²=A²-B=  49-48=1

√(7+1)/2 + √(7-1)/2=  √8/2 + √6/2=  √4 + √3 = 2 + √3

√(7+4√3) -√3= 2 +√3 -√3=  2∈N

Answer 2

[tex] \sqrt{7+4 \sqrt{3} }- \sqrt{3} [/tex]

[tex]7+4 \sqrt{3}=4+2*2 \sqrt{3}+3=(2+ \sqrt{3}) ^{2} [/tex]

$\sqrt{(2+ \sqrt{3}) ^{2} } =2+ \sqrt{3}$

$2+ \sqrt{3}- \sqrt{3}=2$