Question

doresc rezolvarea acestor exercitii prin metoda clasica ci nu cu ajutorul formulei

Answer 1

38.

$a) \: \sqrt{9 + \sqrt{80} } = \\ \sqrt{9 + 4\sqrt{5} } = \\ \sqrt{(2 + \sqrt{5) } {}^{2} } = \\ 2 + \sqrt{5}$

$b) \: \sqrt{10 + \sqrt{84} } = \\ \sqrt{10 + 2 \sqrt{21} } = \\ \sqrt{( \sqrt{3} + \sqrt{7} ) {}^{2} } = \\ \sqrt{3} + \sqrt{7}$

$c) \sqrt{11 + 6 \sqrt{2} } = \\ \sqrt{(3 + \sqrt{2} ){}^{2} } = \\ 3 + \sqrt{2}$

$d) \sqrt{11 - \sqrt{72} } = \\ \sqrt{11 - 6 \sqrt{2} } = \\ \sqrt{(3 - \sqrt{2}) {}^{2} } = \\ 3 - \sqrt{2}$

$e) \sqrt{10 - 4 \sqrt{6} } = \\ \sqrt{(2 - \sqrt{6)} {}^{2} } = \\ \sqrt{6} - 2$

$f) \sqrt{15 - 6 \sqrt{6} } = \\ \sqrt{(3 - \sqrt{6) } {}^{2} } = \\ 3 - \sqrt{6}$

39.

$a) \sqrt{2} + \sqrt{6 - \sqrt{32} } = \\ \sqrt{2} + \sqrt{6 - 4 \sqrt{2} } = \\ \sqrt{2} + \sqrt{(2 - \sqrt{2)} {}^{2} } = \\ \sqrt{2} + 2 - \sqrt{2} = \\ 2$

$b) \sqrt{13 + 4 \sqrt{3} } - \sqrt{12} = \\ \sqrt{(1 + 2 \sqrt{3)} {}^{2} } - 2 \sqrt{3} = \\ 1 + 2 \sqrt{3} - 2 \sqrt{3} = \\ 1$

$c) \sqrt{10 + \sqrt{96} } - \sqrt{15 + \sqrt{216} } = \\ \sqrt{10 + 4 \sqrt{6} } - \sqrt{15 + 6 \sqrt{6} } = \\ \sqrt{(2 + \sqrt{6)} {}^{2} } - \sqrt{(3 + \sqrt{6)} {}^{2} } = \\ 2 + \sqrt{6} - (3 + \sqrt{6} ) = \\ 2 + \sqrt{6} - 3 - \sqrt{6} = \\ - 1$

$d) \sqrt{7 - 4 \sqrt{3} } + \sqrt{4 + 2 \sqrt{3} } = \\ \sqrt{(2 - \sqrt{3)} {}^{2} } + \sqrt{(1 + \sqrt{3)} {}^{2} } = \\ 2 - \sqrt{3} + 1 + \sqrt{3} = \\ 3$