Răspuns:
622
Explicație pas cu pas:
descompunem in factori pe 625, 48 si 243
625 | 5 48 | 2 243 | 3
125 | 5 24 | 2 81 | 3
25 | 5 12 | 2 27 | 3
5 | 5 6 | 2 9 | 3
1 3 | 3 3 | 3
1 1
-> 625 = [tex]5^{4}[/tex]; 48 = 3 * [tex]2^{4}[/tex]; 243 = [tex]3^{5}[/tex] = 3 * [tex]3^{4}[/tex]
stiind ca [tex]\sqrt[4]{a^{4}} = a[/tex], putem face calculele:
-> [tex](\sqrt[4]{5^4} +\sqrt[4]{3} )(5 + \sqrt[4]{3*2^{4}} - \sqrt[4]{3*3^{4}} )[/tex]
-> (5 + [tex]\sqrt[4]{3}[/tex])(5 + [tex]2\sqrt[4]{3} -3\sqrt[4]{3}[/tex])=[tex](5+\sqrt[4]{3} )(5-\sqrt[4]{3} )[/tex]
(a + b)(a - b) = [tex]a^{2} - b^{2}[/tex]
-> [tex]5^{2}-(\sqrt[4]{3}) ^{2}[/tex] = 25 - [tex]\sqrt{3}[/tex] ([tex]\sqrt[4]{3^{2}} = 3^{\frac{2}{4} } = 3^{\frac{1}{2} } = \sqrt{3}[/tex])
[tex]\sqrt{4+\sqrt{12} } = \sqrt{\frac{4+\sqrt{4^{2}-12} }{2} }+ \sqrt{\frac{4-\sqrt{4^{2}-12} }{2} }[/tex]
[tex]\sqrt{\frac{4+\sqrt{4} }{2} }+ \sqrt{\frac{4-\sqrt{4} }{2} }[/tex] = [tex]\sqrt{\frac{4+2}{2} }+ \sqrt{\frac{4-2}{2} } = \sqrt{\frac{6}{2}} +\sqrt{\frac{2}{2} } =\sqrt{3} + 1[/tex]
->[tex]\sqrt{4+\sqrt{12} }+24 = \sqrt{3} +1+24=\sqrt{3} +25[/tex]
->[tex](25-\sqrt{3} )(\sqrt{3} +25)=25^{2} - (\sqrt{3} )^{2} = 625 - 3 = 622[/tex]
