Помогите с 346 и 347

Question 1

Question 2

$(2+ \sqrt{3}) ^{x} +(2- \sqrt{3}) ^{x} =4\\\\2+ \sqrt{3} = \frac{(2+ \sqrt{3})(2- \sqrt{3}) }{2- \sqrt{3} } = \frac{4-3}{2- \sqrt{3} } = \frac{1}{2- \sqrt{3} } \\\\\\ \frac{1}{(2- \sqrt{3}) ^{x} }+(2- \sqrt{3}) ^{x} =4\\\\(2- \sqrt{3}) ^{x} =m\ \textgreater \ 0\\\\ \frac{1}{m}+m-4=0\\\\ m^{2}-4m+1=0\\\\D=(-4) ^{2} -4*1*1=16-4=12=(2 \sqrt{3}) ^{2}\\\\ m_{1} = \frac{4+2 \sqrt{3} }{2}=2+ \sqrt{3}\\\\ m_{2}= \frac{4-2 \sqrt{3} }{2} =2- \sqrt{3} \\\\(2- \sqrt{3}) ^{x} =2+ \sqrt{3}$

$(2- \sqrt{3} ) ^{x}=(2- \sqrt{3}) ^{-1} \\\\ x_{1}=-1\\\\\\(2- \sqrt{3}) ^{x}=2- \sqrt{3} \\\\ x_{2} =1$
Ответ: - 1 ; 1

$(\sqrt{7+ \sqrt{48} } ) ^{x} + (\sqrt{7- \sqrt{48} } ) ^{x}=14\\\\\\ \sqrt{7+ \sqrt{48} }= \frac{( \sqrt{7+ \sqrt{48} })( \sqrt{7- \sqrt{48} }) }{ \sqrt{7- \sqrt{48} }}= \frac{ \sqrt{49-48} }{ \sqrt{7- \sqrt{48} } } = \frac{1}{ \sqrt{7- \sqrt{48} } }$

$\frac{1}{ (\sqrt{7- \sqrt{48} }) ^{x} }+( \sgrt{7- \sqrt{48} }) ^{x} =14\\\\( \sqrt{7- \sqrt{48} }) ^{x}=m\ \textgreater \ 0\\\\\\ \frac{1}{m}+m-14=0\\\\ m^{2} -14m+1=0\\\\D=(-14) ^{2} -4*1*1=196-4=192=8 \sqrt{3} \\\\ m_{1} = \frac{14+8 \sqrt{3} }{2}=7+4 \sqrt{3} \\\\ m_{2} = \frac{14-8 \sqrt{3} }{2}=7-4 \sqrt{3} \\\\\\ (\sqrt{7- \sqrt{48} }) ^{x} =( \sqrt{7-4 \sqrt{3} }) ^{x}=(7-4 \sqrt{3} ) ^{0,5x} \\\\(7-4 \sqrt{3} ) ^{0,5x} =7+4 \sqrt{3} \\\\(7-4 \sqrt{3} ) ^{0,5x}=(7-4 \sqrt{3} )^{-1} \\\\ 0,5x=-1$

$x_{1}=-2\\\\\\(7-4 \sqrt{3} ) ^{0,5x} =7-4 \sqrt{3} \\\\0,5x=1 \\\\ x_{2}=2$