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Условие
Решить уравнение: 2x/(x^2-4x+2) + 3x/(x^2+x+2) = - 5/4
Решение:
![\frac{2x}{x^2-4x+2}+ \frac{3x}{x^2+x+2}=- \frac{5}{4} |\cdot (x^2-4x+2)(x^2+x+2)\cdot 4 \frac{2x}{x^2-4x+2}+ \frac{3x}{x^2+x+2}=- \frac{5}{4} |\cdot (x^2-4x+2)(x^2+x+2)\cdot 4](https://tex.z-dn.net/?f=+%5Cfrac%7B2x%7D%7Bx%5E2-4x%2B2%7D%2B+%5Cfrac%7B3x%7D%7Bx%5E2%2Bx%2B2%7D%3D-+%5Cfrac%7B5%7D%7B4%7D+%7C%5Ccdot+%28x%5E2-4x%2B2%29%28x%5E2%2Bx%2B2%29%5Ccdot+4)
![8x(x^2+x+2)+12x(x^2-4x+2)+5(x^2+x+2)(x^2-4x+2)=0 \\ \\ 1.25(x^2-4x+2)(x^2+x+2)+x(5x^2-10x+10)=0 \\ 1.25((x+ \frac{2}{x} )-4)((x+ \frac{2}{x})+1)+(5(x+ \frac{2}{x})-10)=0 8x(x^2+x+2)+12x(x^2-4x+2)+5(x^2+x+2)(x^2-4x+2)=0 \\ \\ 1.25(x^2-4x+2)(x^2+x+2)+x(5x^2-10x+10)=0 \\ 1.25((x+ \frac{2}{x} )-4)((x+ \frac{2}{x})+1)+(5(x+ \frac{2}{x})-10)=0](https://tex.z-dn.net/?f=8x%28x%5E2%2Bx%2B2%29%2B12x%28x%5E2-4x%2B2%29%2B5%28x%5E2%2Bx%2B2%29%28x%5E2-4x%2B2%29%3D0+%5C%5C++%5C%5C+1.25%28x%5E2-4x%2B2%29%28x%5E2%2Bx%2B2%29%2Bx%285x%5E2-10x%2B10%29%3D0+%5C%5C+1.25%28%28x%2B+%5Cfrac%7B2%7D%7Bx%7D+%29-4%29%28%28x%2B+%5Cfrac%7B2%7D%7Bx%7D%29%2B1%29%2B%285%28x%2B+%5Cfrac%7B2%7D%7Bx%7D%29-10%29%3D0)
Пусть
, тогда получаем
![1.25(t-4)(t+1)+(5t-10)=0 \\ 1.25t^2-3.75t-5+5t-10=0 \\ 1.25t^2+1.25t-15=0 \\ t^2+t-12=0 1.25(t-4)(t+1)+(5t-10)=0 \\ 1.25t^2-3.75t-5+5t-10=0 \\ 1.25t^2+1.25t-15=0 \\ t^2+t-12=0](https://tex.z-dn.net/?f=1.25%28t-4%29%28t%2B1%29%2B%285t-10%29%3D0+%5C%5C+1.25t%5E2-3.75t-5%2B5t-10%3D0+%5C%5C+1.25t%5E2%2B1.25t-15%3D0+%5C%5C+t%5E2%2Bt-12%3D0)
По т. ВИета
![t_1=-4 \\ t_2=3 t_1=-4 \\ t_2=3](https://tex.z-dn.net/?f=t_1%3D-4+%5C%5C+t_2%3D3)
Обратная замена
![x+ \frac{2}{x}=-4|\cdot x \\ x^2+4x+2=0 \\ D=b^2-4ac=16-8=8; \sqrt{D} =2 \sqrt{2} \\ x_1_,_2=-2\pm \sqrt{2} x+ \frac{2}{x}=-4|\cdot x \\ x^2+4x+2=0 \\ D=b^2-4ac=16-8=8; \sqrt{D} =2 \sqrt{2} \\ x_1_,_2=-2\pm \sqrt{2}](https://tex.z-dn.net/?f=x%2B+%5Cfrac%7B2%7D%7Bx%7D%3D-4%7C%5Ccdot+x+%5C%5C+x%5E2%2B4x%2B2%3D0+%5C%5C+D%3Db%5E2-4ac%3D16-8%3D8%3B+%5Csqrt%7BD%7D+%3D2+%5Csqrt%7B2%7D++%5C%5C+x_1_%2C_2%3D-2%5Cpm++%5Csqrt%7B2%7D+)
![x+ \frac{2}{x}=3|\cdot x \\ x^2-3x+2=0 x+ \frac{2}{x}=3|\cdot x \\ x^2-3x+2=0](https://tex.z-dn.net/?f=x%2B+%5Cfrac%7B2%7D%7Bx%7D%3D3%7C%5Ccdot+x+%5C%5C+x%5E2-3x%2B2%3D0)
По т. ВИета
![x_3=1 \\ x_4=2 x_3=1 \\ x_4=2](https://tex.z-dn.net/?f=x_3%3D1+%5C%5C+x_4%3D2)
Решить уравнение: 2x/(x^2-4x+2) + 3x/(x^2+x+2) = - 5/4
Решение:
Пусть
По т. ВИета
Обратная замена
По т. ВИета
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