Question

$frac{3x}{x^{2}-2x+3}+frac{6x}{x^{2}-3x+3}=14$ помогите решить

Answer 1

$frac{3x}{x^2-2x+3}+frac{6x}{x^2-3x+3}=14;\\ x^2-2x+3 neq 0; x^2-3x+3 neq 0;\\ (x-3)(x+1) neq 0;x^2-3x+3=(x-1.5)^2+0.75>0;\\ x neq 3; x neq -1;\\ frac{3}{x-2+frac{3}{x}}+frac{6}{x-3+frac{3}{x}}=14;$

$y=x+frac{3}{x}-2.5;\\ frac{3}{y+0.5}+frac{6}{y-0.5}=14;\\ 3(y-0.5)+6(y+0.5)=14(y+0.5)(y-0.5);\\ 3y-1.5+6y+3=14y^2-3.5;\\ 14y^2-9y-5=0;\\D=(-9)^2-4*14*(-5)=361=19^2;\\ y_1=frac{9-19}{2*14}=-frac{10}{28}=-frac{5}{14};\\ y_2=frac{9+19}{2*14}=frac{28}{28}=1;$

$x+frac{3}{x}-2.5=-frac{5}{14};\\14x^2-35x+42+5x=0;\\14x^2-30x+42=0;\\7x^2-15x+21=0;\\D=(-15)^2-4*7*21=-363<0;\\x+frac{3}{x}-2.5=1;\\x^2-3.5x+3=0;\\D=(-3.5)^2-4*1*3=0.25;\\x_1=frac{3.5-0.25}{2*1}=1.625;\\x_2=frac{3.5+0.25}{2*1}=1.875$