Уравнение 1/x^2-2x+2 + 2/x^2-2x+3=6/x^2-2x+4

[Гост]

1/x^2-2x+2 + 2/x^2-2x+3=6/x^2-2x+4

Благодаря предваритено

[ammornil]

[tex]\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4} \\
\cyr{DM}:\hspace{4} \left|x^2-2x+2 \ne 0 \\ x^2-2x+3 \ne 0 \\ x^2-2x+4 \ne 0 \right, \hspace{4} \Rightarrow \left| D<0 \\ D<0 \\ D<0 \right, \hspace{4} \Rightarrow \forall x \in R \\
\vspace{10}\\
\cyr{polagame} \hspace{12} x^2-2x=u \\
\vspace{10}\\
\frac{1}{u+2}+\frac{2}{u+3}=\frac{6}{u+4} \\
(u+3).(u+4)+2.(u+2).(u+4)-6.(u+2).(u+3)=0 \\
u^2+4u+3u+12+2u^2+8u+4u+16-6u^2-18u-12u-36=0 \\
u^2+2u^2-6u^2+4u+3u+8u+4u-18u-12u+12+16-36=0 \\
-3u^2-11u-8=0 \hspace{8} \Leftrightarrow \hspace{8} 3u^2+11u+8=0 \\
D=11^2-4.3.8=121-96=25 \hspace{12} \sqrt{D}=5 \\
u_{_{1,2}}=\frac{-11 \pm 5}{6} \hspace{12} \left{u_{_{1}}=-\frac{8}{3} \\ u_{_{2}}=-1 \right, \\
\vspace{10}\\
x^2-2x=-\frac{8}{3} \hspace{60} \cup \hspace{60} x^2-2x=-1 \\
3x^2-6x+8=0 \hspace{45} \cup \hspace{60} x^2-2x+1=0 \\
D<0 \hspace{120} \cup \hspace{60} (x-1)^2=0 \\
x \in \oslash \hspace{120} \cup \hspace{60} x=1\\
\vspace{25} \\
\cyr{OTGOVOR}: x=1[/tex]