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December 2021 1 5 Report

----Решите уравнение----

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Universalka

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$\sqrt{x-1+\sqrt{x+2}}=3\\\\(\sqrt{x-1+\sqrt{x+2}})^{2} =3^{2}\\\\x-1+\sqrt{x+2}=9\\\\\sqrt{x+2}=10-x\\\\\left \{ {{(\sqrt{x+2})^{2} =(10-x)^{2} } \atop {10-x\geq0 }} \right.\\\\\left \{ {{x+2=100-20x+x^{2} } \atop {-2\leq x\leq10 }} \right.\\\\\left \{ {{x^{2}-21x+98=0 } \atop {-2\leq x \leq10 }} \right. \\\\x^{2}-21x+98=0\\\\D=(-21)^{2}-4*98=441-392= 49=7^{2}\\\\x_{1}=\frac{21-7}{2}=7\\\\x_{2}=\frac{21+7}{2}=14-neyd\\\\Proverka:\\\\\sqrt{7-1+\sqrt{7+2} }=3$

$\sqrt{6+\sqrt{9} } =3\\\\\sqrt{6+3}=3\\\\\sqrt{9}=3\\\\3=3- \ verno\\\\Otvet:\boxed7$

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