Помогите решить пожалуйста 3 примера)
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1) \frac{3x^{2} -5x-2}{2-x} = 0
ОДЗ:
2-x≠0 ; x≠2. \frac{2y-5}{y+5} =  \frac{3(y+7)}{2y-1} ; \\ 3(y+7)(y+5) = (2y-5)(2y-1); \\ 3(y^{2}+12y+35) = 4y^{2} -12y+5; \\ 3y^{2} + 36y + 105 = 4y^{2}  - 12y + 5; \\  y^{2} - 48y - 100 = 0; \\  \sqrt{D} =  \sqrt{2304+400} = \sqrt{2704} = 52 \\ y_{1} =  \frac{48-52}{2} = -2; \\ y_{2} =  \frac{48+52}{2} = 50;

3x^{2}-5x-2 = 0 \\ \sqrt{D} = \sqrt{25+24}= \sqrt{49} = 7 \\ x_{1} = \frac{5-7}{6} = - \frac{1}{3} \\ x_{2} = \frac{5+7}{6} =2
Так как x не может равняться 2, тогда уравнение имеет один корень - (- \frac{1}{3} ).

2)  \frac{2y-5}{y+5} =  \frac{3y+21}{2y-1};
ОДЗ
 \left \{ {{y+5 \neq 0} \atop {2y-1 \neq 0}} \right.  \\  \left \{ {{y \neq -5} \atop {y \neq 0,5}} \right.

3)  \frac{7}{x-3} +1 = \frac{18}{x^{2}-6x+9};
ОДЗ
 \left \{ {{x-3 \neq 0} \atop {x^{2}-6+9 \neq 0}} \right.  \\  \left \{ {{x \neq 3} \atop {x \neq 3}} \right.
 \frac{4+x}{x-3} =  \frac{18}{(x-3)(x-3)} ; \\  \frac{4+x}{x-3} -  \frac{18}{(x-3)(x-3)} = 0; \\  \frac{(x+4)(x-3)-18}{(x-3)(x-3)}  = 0; \\  x^{2} +x-30 = 0;  \\ \sqrt{D} =  \sqrt{1+120} =  \sqrt{121} = 11; \\  x_{1} =  \frac{-1-11}{2} = -6; \\  x_{2} =  \frac{-1+11}{2} = 5.
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