Задача-Виет

[Гост]

Ако някой знае, моля да напише!

[ammornil]

[tex]kx^{2}-(k+4)x+k+5=0[/tex]
[tex]\begin{array}{|l} x_{1}+x_{2}=-\Large{\frac{-(k+4)}{k}} \\ x_{1}.x_{2}=\Large{\frac{k+5}{k}} \end{array} \Leftrightarrow \begin{array}{|l} x_{1}+x_{2}=\Large{\frac{k+4}{k}} \\ x_{1}.x_{2}=\Large{\frac{k+5}{k}} \end{array}[/tex]

Ето няколко полезни трансформации, които се ползват за решаване на дадените проблеми
(A) [tex](x_{1}+x_{2})^{2}=x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2} \Rightarrow x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}x_{2} \Leftrightarrow[/tex] $$ x_{1}^{2}+x_{2}^{2}=\left(\frac{k+4}{k} \right)^{2}-2\frac{k+5}{k}$$

(Б) [tex]x_{1}=\frac{-b+\sqrt{D}}{2}, x_{2}=\frac{-b-\sqrt{D}}{2}, x_{1}-x_{2}=\frac{-b+\sqrt{D}}{2}-\frac{-b-\sqrt{D}}{2}=\sqrt{D}[/tex]
$$ D=(k+4)^{2}-4k(k+5) $$

(В) [tex](\sqrt{x_{1}}+\sqrt{x_{1}})^{2}=(\sqrt{x_{1}})^{2}+2\sqrt{x_{1}}\sqrt{x_{2}}+(\sqrt{x_{2}})^{2} \Rightarrow (\sqrt{x_{1}}+\sqrt{x_{1}})^{2}=x_{1}+2\sqrt{x_{1}x_{2}}+x_{2}[/tex] $$ \sqrt{x_{1}}+\sqrt{x_{1}} = \sqrt{(x_{1}+x_{2})+2\sqrt{x_{1}x_{2}}} $$

[Евва]

Ето още една идея :
А=[tex]x_{1 } ^{2 }[/tex][tex]x_{2 } ^{3 }[/tex] +[tex]x_{1 } ^{3 }[/tex][tex]x_{2 } ^{2 }[/tex]=

=[tex]x_{1 } ^{2 }[/tex][tex]x_{2 } ^{2 }[/tex]([tex]x_{2 }[/tex]+[tex]x_{1 }[/tex]) =

=[tex]( x_{1 } x_{2 } )^{2 }[/tex]([tex]x_{1 } +x_{2}[/tex])=

=([tex]\frac{к+5}{к}) ^{2 }[/tex].[tex]\frac{к+4}{к}[/tex] =[tex]\frac{ к^{2 }+10к+25 }{ к^{2 } }[/tex].[tex]\frac{к+4}{к}[/tex]=

=[tex]\frac{ к^{3 } +4 к^{2 }+10 к^{2 }+40к+25к+100 }{ к^{3 } }[/tex]=

=[tex]\frac{ к^{3 } +14 к^{2 } +65к+100}{ к^{3 } }[/tex]

[mail_dinko]

[tex]\begin{array}{|l} x_{1}+x_{2}=\Large{\frac{k+4}{k}} \\ x_{1}.x_{2}=\Large{\frac{k+5}{k}} \end{array}[/tex]
[tex]A=x_1^2 x_2^3+x_1^3 x_2^2=(x_1x_2)^2 (x_1+x_2)= \frac {(k+5)^2}{k^2}.\frac {k+4}{k}= \frac {k^3+14k^2+65k+100}{k^3}[/tex]

[tex]B=x_1-x_2\Rightarrow B^2=x_1^2+x_2^2-2x_1x_2 = (x_1+x_2)^2-4x_1x_2 =\frac {(k+4)^2}{k^2}- 4. \frac {k+5}{k}[/tex]
[tex]B=\sqrt {\frac {k^2+8k+16-4k^2-20k}{k^2}} = \frac {\sqrt {-3k^2-12k+16}}{k}[/tex]

[tex]C = \sqrt {x_1}+ \sqrt {x_2}\Rightarrow C^2 = x_1+x_2+2 \sqrt {x_1x_2}= \frac {k+4}{k}+2 \sqrt {\frac {k+5}{k}}[/tex]
[tex]C= \sqrt { \frac {k+4}{k}+2 \sqrt {\frac {k+5}{k}}}[/tex]