Question

Номер 9, пожалуйста, с решением 40 баллов

Answer 1

$x^{2} +5x-3=0\\\\Teorema \ Vieta:\\\\x_{1}+x_{2} =-5\\\\x_{1}\cdot x_{2} =-3\\\\\\1)x_{1}+x_{2} +x_{1}\cdot x_{2}=-5-3=\boxed{-8}\\\\\\2)\frac{1}{x_{1} }+\frac{1}{x_{2} }=\frac{x_{1}+x_{2} }{x_{1}\cdot x_{2} }=\frac{-5}{-3}=\boxed{1\frac{2}{3}}\\\\\\3)x_{1}^{2} \cdot x_{2}+x_{2}^{2}\cdot x_{1}=x_{1}\cdot x_{2}(x_{1}+x_{2})=-3\cdot (-5) =\boxed{15} \\\\\\4)x_{1}^{2} +x_{2}^{2}=(x_{1}+x_{2})^{2} -2\cdot x_{1}\cdot x_{2} =(-5)^{2}-2\cdot (-3)=25+6=\boxed{31}$

Answer 2

Объяснение:

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