Question

Fie numarul a =4(√3+√2)²+2(√2+2√3)(√3-2√2)+(√3-√2)², aratati ca a este numar natural impar

Answer 1

Explicație pas cu pas:

$a = 4( \sqrt{3} + \sqrt{2}) ^{2} + 2( \sqrt{2} + 2 \sqrt{3})( \sqrt{3} - 2 \sqrt{2}) + ( \sqrt{3} - \sqrt{2}) ^{2} \\ \\ a = 4( \sqrt{3} ^{2} + 2 \sqrt{3} \sqrt{2} + \sqrt{2} ^{2}) + 2( \sqrt{2} + 2 \sqrt{3})( \sqrt{3} - 2 \sqrt{2}) + ( \sqrt{3} - \sqrt{2}) ^{2} \\ \\ a = 12 + 8 \sqrt{6} + 8 + 2( \sqrt{2} + 2 \sqrt{3})( \sqrt{3} - 2 \sqrt{2}) + ( \sqrt{3} - \sqrt{2}) ^{2} \\ \\ a = 12 + 8 \sqrt{6} + 8 + (2 \sqrt{2} + 4 \sqrt{3})( \sqrt{3} - 2 \sqrt{2}) + ( \sqrt{3} - \sqrt{2}) ^{2} \\ \\ a = 12 + 8 \sqrt{6} + 8 + 2 \sqrt{6} - 8 + 12 - 8 \sqrt{6} + ( \sqrt{3} - \sqrt{2}) ^{2} \\ \\ a = 12 + 8 \sqrt{6} + 8 + 2 \sqrt{6} - 8 + 12 - 8 \sqrt{6} + \sqrt{3} ^{2} - 2 \sqrt{3} \sqrt{2} + \sqrt{2} ^{2} \\ \\ a = 12 + 8 \sqrt{6} + 8 + 2 \sqrt{6} - 8 + 12 - 8 \sqrt{6} + 3 - 2 \sqrt{3} \sqrt{2} + 2 \\ \\ a = 12 + 8 \sqrt{6} + 8 + 2 \sqrt{6} - 8 + 12 - 8 \sqrt{6} + 3 - 2 \sqrt{3 \times 2} + 2 \\ \\ a = 12 + 8 \sqrt{6} + 8 + 2 \sqrt{6} - 8 + 12 - 8 \sqrt{6} + 3 - 2 \sqrt{6} + 2 \\ \\ a = (12 + 8 - 8 + 12 + 3 + 2) + (8 \sqrt{6} + 2 \sqrt{6} - 8 \sqrt{6} - 2 \sqrt{6}) \\ \\ a = 29$