Question

$\\if \: x + 1 \div x = 4 \: \\ and \\ \: {x}^{2} + 1 \div {x}^{2} = 10 \\ then \\ find \: the \: value \: of \: x - 1 \div x.$here is your question

Answer 1

$( \frac{x + 1}{x} ) = 4 \\ x + 1 = 4x \\ x = \frac{1}{3} \\ \frac{x {}^{2} + 1 }{ {x}^{2} } = 10 \\ we \: have \: x = \frac{1}{3} \: lets \: check \: it \\ \frac{ { \frac{1}{3} }^{2} + 1}{ { \frac{1}{3} }^{2} } = 10 \\ so \\ \frac{x - 1}{x} = \frac{( \frac{1}{3}) - 1 }{ \frac{1}{3} } = - 2$
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Answer 2

the value of (x-1/x) = 2.83
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