Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!! 15 балов!! ЗАРАНЕЕ СПАСИБО ❤️​

Answer 1

$4x^{2}-5x-13=0|:4\\\\x^{2}-\frac{5}{4}x-\frac{13}{4}=0\\\\x_{1} +x_{2}= \frac{5}{4} \\\\x_{1} *x_{2}=-\frac{13}{4}\\\\\\1)x_{1} *x_{2}-2x_{1} -2x_{2}=x_{1} *x_{2}-2(x_{1} +x_{2})=-\frac{13}{4} -2*\frac{5}{4}=\\\\=-\frac{13}{4}-\frac{10}{4}=\boxed{-\frac{23}{4}}\\\\\\2)x_{1} ^{2} +x_{2}^{2}=(x_{1} +x_{2})^{2}-2x_{1}*x_{2}= (\frac{5}{4})^{2}-2*(-\frac{13}{4}=\frac{25}{16}+\frac{26}{4}=\\\\=\frac{25}{16}+\frac{104}{16}=\boxed{\frac{129}{16}}$

$3)x_{1}^{3}+x_{2}^{3} =(x_{1}+x_{2})[(x_{1}+x_{2})^{2}-3x_{1}*x_{2}]=\frac{5}{4}*[(\frac{5}{4})^{2}-3*(-\frac{13}{4})]= \\\\=\frac{5}{4}*(\frac{25}{16} +\frac{39}{4})=\frac{5}{4}*(\frac{25}{16}+\frac{156}{16})=\frac{5}{4}*\frac{181}{16} =\boxed{\frac{905}{64}}\\\\\\4)x_{1}^{2}*x_{2}+x_{1}*x_{2}^{2} =x_{1}*x_{2}(x_{1}+x_{2} )=-\frac{13}{4}*\frac{5}{4}=\boxed{-\frac{65}{16}}$

$5)(x_{1}-x_{2})^{2}=(x_{1}+x_{2})^{2}-4(x_{1}*x_{2}) =(\frac{5}{4})^{2}-4*(-\frac{13}{4} )=\\\\=\frac{25}{16}+\frac{208}{16}=\boxed{\frac{233}{16} }$