Question

помогите решить

$12x {}^{2} + 3x = 0$
$x {}^{2} - 12 = 0$

$\frac{x}{20 - x} = \frac{ 1}{x}$
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Answer 1

Ответ:

Объяснение:

$12{x}^{2} + 3x = 0\\3x(4x+1)=0\\x=o; \\4x+1=0; x=-\frac{1}{4} \\....................\\ {x}^{2} - 12 = 0\\ {x}^{2} =12\\x=\sqrt{12} =2\sqrt{3} ; x=-\sqrt{12}=-2\sqrt{3} ; \\......................\\\frac{x}{20 - x} = \frac{ 1}{x} ; x\neq 20\\x^{2} =20-x\\x^{2} +x-20=0; D=1+80=81>0;\\x_{1} =\frac{-1+9}{2} =4;\\x_{2} =\frac{-1-9}{2} =-5;$