3 х + 4 x^2 ________ = 0 x-1x^2 -4x______=0 x+3 9-x^2_____=0 x+336x^2-1______=0 1-6x

$\frac{3x+4x^2}{x-1} =0 \\ \frac{x(3+4x)}{x-1}=0 \\ \left \{ {{x(3+4x)=0} \atop {x-1 \neq 0}} \right. \\ \left \{ {{x=0, x= - \frac{3}4} } \atop {x \neq 1}} \right.$
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$\frac{x^2-4x}{x+3} =0 \\ \frac{x(x-4)}{x+3}=0 \\ \left \{ {{x(x-4)=0} \atop {x+3 \neq 0}} \right. \\ \left \{ {{x=0, x=4} \atop {x \neq -3}} \right.$
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$\frac{9-x^2}{x+3}=0 \\ \frac{(3-x)(3+x)}{x+3} =0 \\ 3-x=0 \\ x=-3$
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$\frac{36x^2-1}{1-6x}=0 \\ \frac{(6x-1)(6x+1)}{1-6x}=0 \\ -\frac{(6x-1)(6x+1)}{6x-1}=0 \\ -6x+1=0 \\ 6x-1=0 \\ 6x=1 \\ x= \frac{1}{6}$
Надеюсь я правильно вас поняла.