Question

$(\sqrt{4+\sqrt{15} }) ^{x}+ (\sqrt{4-\sqrt{15} }) ^{x}=8$

Answer 1

Ответ$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\dddddd \displaystyle \large \boldsymbol{:}$

Объяснение:

$\large \boldsymbol {} ( (\sqrt{4+\sqrt{15} } )^x+(\sqrt{4-\sqrt{15} } )^x)^2 =8^2 \\\\(4+\sqrt{15} )^x+2\cdot \sqrt{\underbrace{((4^2-15)}_1})^x +(4-\sqrt{15} )^x=64$

Преобразуем :

$\large \boldsymbol {} (4-\sqrt{15}) \cdot \dfrac{4+\sqrt{15} }{4+\sqrt{15} } =\dfrac{1}{4+\sqrt{15} }$

$\large \boldsymbol {} (4+\sqrt{15} )^x+2 +\bigg(\dfrac{1}{4+\sqrt{15}} \bigg)^x=64$

Сделаем замену:

$\large \boldsymbol {} t=(4+\sqrt{15} )^x \ ; \ \dfrac{1}{t} = \bigg(\dfrac{1}{4+\sqrt{15}} \bigg)^x$

$\displaystyle \large \boldsymbol{} t+\frac{1}{t} -62 = 0 \ \ |\times t \\\\t^2-62t+1=0 \\\\D=62^2-2^2=(62-2)(62+2)=3840 \\\\t_1=\frac{62+16\sqrt{15} }{2} =31+8\sqrt{15} \\\\\\t_2=\frac{62-16\sqrt{15} }{2} =31-8\sqrt{15}$

$\large \boldsymbol \displaystyle 1) \ (4+\sqrt{15} )^x=(4+\sqrt{15} )^2 \\\\\boxed{x=2} \\\\ 2) \ (4+\sqrt{15} )^x=(4-\sqrt{15} )^2 \\\\(4+\sqrt{15} )^x=\bigg(\dfrac{1}{4+\sqrt{15} } \bigg)^2 \\\\\boxed{x=-2}$

Answer 2

Ответ:

x={-2;2}

Объяснение:

/////////////////////////