Question

ajutati-ma va rog dau coroana va rog​

Answer 1

Răspuns:

$a) {x}^{2} - 7x + 6 = {( \sqrt{3} - \sqrt{2}) }^{2} - 7 \times ( \sqrt{3} - \sqrt{2} ) + 6 = 3 - 2 \times \sqrt{3} \times \sqrt{2} + 2 - 7 \sqrt{3} + 7 \sqrt{2} + 6 = 3 - 2 \sqrt{6} + 2 - 7 \sqrt{3} + 7 \sqrt{2} = 5 - 2 \sqrt{6} - 7 \sqrt{3} + 7 \sqrt{2}$

$b) {(5 - \sqrt{2}) }^{2} - 7 \times (5 - \sqrt{2} ) + 6 = 25 - 10 \sqrt{2} + 2 - 35 + 7 \sqrt{2} + 6 = - 2 - 10 \sqrt{2}$

$c)(2 - \sqrt{3} )^{2} - 7 \times (2 - \sqrt{3} ) + 6 = 4 - 4 \sqrt{3} + 3 - 14 + 7 \sqrt{3} + 6 = - 1 + 3 \sqrt{3}$

$( \frac{1}{2} + \sqrt{7} ) ^{2} - 7 \times ( \frac{1}{2} + \sqrt{7}) + 6 = \frac{1}{4} + \sqrt{7} + 7 - \frac{7}{2} - 7 \sqrt{7} + 6 = \frac{1}{4} + 7 - \frac{7}{2} - 6 \sqrt{7} = \frac{1 + 28 - 14 - 42 \sqrt{7} }{7} = \frac{ - 41 - 42 \sqrt{7} }{7}$