Question

9. (97-6-68) Найти x из условия, что
$\tan( \alpha ) = \frac{3 + \sqrt{x} }{2}$
$\tan( \beta ) = \frac{3 - \sqrt{x} }{2}$
$\alpha + \beta = \frac{\pi}{4}$

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Answer 1

$\displaystyle\bf\\tg\alpha =\frac{3+\sqrt{x} }{2} \\\\\\tg\beta =\frac{3-x^{2} }{2} \\\\\\\alpha +\beta =\frac{\pi }{4} \\\\\\tg(\alpha +\beta )=\frac{tg\alpha +tg\beta }{1-tg\alpha \cdot tg\beta } \\\\\\tg\frac{\pi }{4} =\frac{\frac{3+\sqrt{x} }{2} +\frac{3-\sqrt{x} }{2} }{1-\frac{3+\sqrt{x} }{2} \cdot\frac{3-\sqrt{x} }{2} } \\\\\\1=\frac{3}{1-\frac{9-x}{4} }$

$\displaystyle\bf\\1=\frac{3}{\frac{4-9+x}{4} } \\\\\\\frac{12}{x-5} =1\\\\\\x-5=12\\\\\\x=17$

Answer 2

Ответ:

$tg\alpha =\dfrac{3+\sqrt{x}}{2}\ \ ,\ \ tg\beta =\dfrac{3-\sqrt{x}}{2}$

$\alpha +\beta =\dfrac{\pi}{4}\ \ ,\ \ \ tg(\alpha +\beta )=tg\dfrac{\pi}{4}=1\\\\\\tg(\alpha +\beta )=\dfrac{tg\alpha +tg\beta }{1-tg\alpha \cdot tg\beta }=\dfrac{\dfrac{3+\sqrt{x}}{2}+\dfrac{3-\sqrt{x}}{2}}{1-\dfrac{3+\sqrt{x} }{2}\cdot \dfrac{3-\sqrt{x}}{2}}=\dfrac{3+\sqrt{x}+3-\sqrt{x}}{2\cdot \dfrac{4-(3+\sqrt{x})(3-\sqrt{x})}{4}}=\\\\\\=\dfrac{6\cdot 2}{4-(9-x)}=\dfrac{12}{x-5}=1\ \ \ \Rightarrow \ \ \ 12=x-5\ \ \ (x\ne 5)\\\\\\x=12+5\ \ ,\ \ \boxed{x=17\ }$