Question

Решить уравнение
$\frac{x}{ {x}^{2} - 16x } + \frac{x - 1}{x + 4 } = 1$
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Answer 1

$\frac{x}{x^{2}-16x } + \frac{x-1}{x+4} =1 \\\\\frac{x}{x^{2-16x} } + \frac{x-1}{x+4} =1, x\neq 0, x\neq, x\neq -4\\\frac{x}{x*(x-16)} + \frac{x-1}{x+4} =1\\\\\frac{1}{x-16} + \frac{x-1}{x+4}=1\\\\\frac{x+4+(x-16)*(x-1)}{(x-16)*(x+4)}=1\\\frac{x+4+x^{2 }-x-16x+16}{(x-16)*(x+4)}=1\\\\\frac{20+x^{2}-16x}{(x-16)*(x+4)} \\\\20+x^{2} -16x=(x-16)*(x+4)\\20+x^{2}-16x=x^{2}+4x-16x-64\\20=4x-64\\-4x=-64-20\\-4x=-84\\x=21, x\neq0, x\neq16, x\neq-4\\x=21$