Question

Daca a= (2+rad din 3)^2 +rad din 6(2 supra rad din 3-4 supra rad din 2) + (1-rad din 2)^2
Va rog mult!!!!

Answer 1

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Answer 2

$a = {(2 + \sqrt{3}) }^{2} + \sqrt{6}( \frac{2}{ \sqrt{3} } - \frac{4}{ \sqrt{2} }) + {(1 - \sqrt{2}) }^{2} \\ \\ a = {(2 + \sqrt{3}) }^{2} + \sqrt{6}( \frac{2 \sqrt{3} }{3} - \frac{4 \sqrt{2} }{2}) + {(1 - \sqrt{2}) }^{2} \\ \\ a = {2}^{2} + 2 \times 2 \sqrt{3} + { \sqrt{3} }^{2} + 2 \sqrt{2} - 4 \sqrt{3} + 1 - 2 \sqrt{2} + { \sqrt{2} }^{2} \\ \\ a = {2}^{2} + 2 \times 2 \sqrt{3} + 3 + 2 \sqrt{2} - 4 \sqrt{3} + 1 - 2 \sqrt{2} + 2 \\ \\ a = {2}^{2} + 4 \sqrt{3} + 3 + 2 \sqrt{2} - 4 \sqrt{3} + 1 - 2 \sqrt{2} + 2 \\ \\ a = {2}^{2} + (4 \sqrt{3} - 4 \sqrt{3}) + (3 + 1 + 2) + (2 \sqrt{2} - 2 \sqrt{2}) \\ \\ a = {2}^{2} + 6 \\ \\ a = 4 + 6 \\ \\ a = 10$