9. (97-6-68) Найти x из условия, что[tex] \tan( \alpha ) = \frac{3 + \sqrt{x} }{2} [/tex][tex] \tan( \beta ) = \frac{3 - \sqrt{x} }{2} [/tex][tex] \alpha + \beta = \frac{\pi}{4} [/tex]. Created by alinaabasova75. algebra-ru

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

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

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