Question

Решите пожалуйста 1 варинат на листочке

Answer 1

№1. D=4+48=52
$x_{1}= \frac{2+ \sqrt{52} }{6}; x_{2}= \frac{2- \sqrt{52} }{6}; a) x_{1} + x_{2} = \frac{2+ \sqrt{52}+2- \sqrt{52} }{6}= \frac{4}{6}= \frac{2}{3}$
б) $x_{1} x_{2}= \frac{2+ \sqrt{52} }{6} * \frac{2- \sqrt{52} }{6}= \frac{(2+ \sqrt{52})(2- \sqrt{52})} {36}= \frac{4-52}{36}= \frac{-48}{36}= \frac{-4}{3}=-1 \frac{1}{3}$
в) $\frac{1}{ x_{1} } + \frac{1}{ x_{2} }= \frac{6}{2+ \sqrt{52} }+ \frac{6}{2- \sqrt{52} } =$$\frac{12-6 \sqrt{52} }{(2+ \sqrt{52})(2- \sqrt{52}) } }+ \frac{12+6 \sqrt{52} }{(2+ \sqrt{52})(2- \sqrt{52}) } = \frac{12-6 \sqrt{52}+12+6 \sqrt{52}}{4-52}= \frac{24}{-48}= -\frac{1}{2}$
г) $x_{1} ^{2} x_{2} + x_{1} x_{2} ^{2} = x_{1} x_{2}( x_{1} + x_{2} ) = \frac{-4}{3} * \frac{2}{3}= \frac{-8}{9}$
№2. $x_{1}= \frac{-3}{5} ; x_{2} = \frac{1}{2} ;$
$\left \{ {{ x_{1} +x_{2}=-p} \atop { x_{1} x_{2} =q}} \right. \left \{ {{ \frac{-3}{5} + \frac{1}{2} =-p} \atop { \frac{-3}{5} * \frac{1}{2} =q}} \right. \left \{ {{ \frac{-6}{10} + \frac{5}{10} =-p} \atop { \frac{-3*1}{5*2} =q}} \right. \left \{ {{ \frac{-1}{10} =-p} \atop { \frac{-3}{10} =q}} \right. \left \{ {{ -0.1 =-p} \atop { -0.3 =q}} \right. \left \{ {{ 0.1 =p} \atop { -0.3 =q}} \right.$
$x^{2} +px+q=0$
$x^{2} +0.1x-0.3=0$
№3. D=381*381+4*5*386=145161+7720=152881
$x_{1}= \frac{-381+ \sqrt{152881} }{10} = \frac{-381+391}{10} = \frac{10}{10} =1$
$x_{2} = \frac{-381- \sqrt{152881} }{10} = \frac{-381-391}{10}= \frac{-772}{10}= -77.2$