Question

$\frac{4}{x} + 5 = \frac{1}{x {}^{2} }$
помогите пожалуйста ​

Answer 1

Відповідь:

$x_{1}=0.2\\x_{2}=-1$

Пояснення:

x≠0

$4x+5x^{2}=1\\5x^{2}+4x-1=0$

Найдем дискриминант:

$D=16+20=36$

$x_{1} =\frac{-4+6}{10} \\x_{2} =\frac{-4-6}{10} \\x_{1}=0.2\\x_{2}=-1$

Answer 2

$\frac{4}{x}+5=\frac{1}{x^{2} }, x\neq  0\\\\\frac{4}{x}+5-\frac{1}{x^{2} }=0\\\\\frac{4x+5x^{2}-1 }{x^{2} } =0\\\\4x+5x^{2} -1=0\\\\5x^{2} +4x-1=0\\\\5x^{2} +5x-x-1=0\\\\5x*(x+1)-(x+1)=0\\\\(x+1)(5x-1)=0\\\\x+1=0\\\\5x-1=0\\\\x=-1\\\\x=\frac{1}{5}\\\\x_{1} =-1, x_{2}=\frac{1}{5}$