Question

19. Докажите тождество: 1) sqrt(9 - 2sqrt(14)) = sqrt(7) - sqrt(2) 3) sqrt(7 + 4sqrt(3)) + sqrt(7 - 4sqrt(3)) = 4 2) sqrt(6sqrt(2) + 11) = sqrt(2) + 3 4) √8+2√7 + √8-2√7 N 27.​

Answer 1

$\displaystyle\bf\\1)\\\\\sqrt{9-2\sqrt{14} } =\sqrt{7} -\sqrt{2} \\\\\\ \sqrt{9-2\sqrt{14} } =\sqrt{7-2\sqrt{14}+2 } =\\\\\\=\sqrt{(\sqrt{7} )^{2} -2\cdot\sqrt{7}\cdot\sqrt{2} +(\sqrt{2} )^{2} } =\\\\\\=\sqrt{(\sqrt{7} -\sqrt{2} )^{2} } =\sqrt{7} -\sqrt{2} \\\\2)\\\\\sqrt{6\sqrt{2}+11 } =\sqrt{2} +3 \\\\\\ \sqrt{6\sqrt{2} +11} =\sqrt{2+6\sqrt{2}+9 } =\\\\\\=\sqrt{(\sqrt{2} )^{2} +2\cdot\sqrt{2}\cdot 3 +3^{2} } =\\\\\\=\sqrt{(\sqrt{2} +3)^{2} } =\sqrt{2} +3$

$\displaystyle\bf\\3)\\\\\sqrt{7+4\sqrt{3} } +\sqrt{7-4\sqrt{3} }=27 \\\\\\ \sqrt{7+4\sqrt{3} } +\sqrt{7-4\sqrt{3} } =\\\\\\=\sqrt{2^{2} +2\cdot 2\cdot\sqrt{3} +(\sqrt{3}) ^{2} } +\sqrt{2^{2} -2\cdot 2\cdot\sqrt{3} +(\sqrt{3} )^{2} } =\\\\\\=\sqrt{(2+\sqrt{3} )^{2} } +\sqrt{(2-\sqrt{3} )^{2} } =2+\sqrt{3} +2-\sqrt{3} =4\\\\4)$

$\displaystyle\bf\\\sqrt{8+2\sqrt{7} } +\sqrt{8-2\sqrt{7} }=2\sqrt{7} \\\\\\ \sqrt{7+2\sqrt{7} +1} +\sqrt{7-2\sqrt{7}+1 } =\\\\\\=\sqrt{(\sqrt{7} )^{2} +2\cdot 1\cdot\sqrt{7} +1^{2} } +\sqrt{(\sqrt{7}) ^{2} -2\cdot 1\cdot\sqrt{7} +1^{2} } =\\\\\\=\sqrt{(\sqrt{7} +1)^{2} } +\sqrt{(\sqrt{7} -1)^{2} } =\sqrt{7} +1+\sqrt{7}-1=2\sqrt{7}$