Question

solve 1/x -1/x-2 = 3 ; x,y ≠ 0​

Answer 1

Step-by-step explanation:

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Answer 2

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Questions :

$\rm solve \to \: \\ \bigstar \bf\frac{1}{x} - \frac{1}{x - 2} = 3$

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{w}\purple{E}\orange{R}}}}$

$\\ \implies \sf \: \frac{1}{x} - \frac{1}{x - 2} = 3$

$\\ \implies \large \sf \: \frac{(x - 2) - x}{(x - 2)(x)} = 3$

$\\ \implies \large \sf \: \frac{ \cancel{x} \: - 2 - \cancel{x}}{ {x}^{2} - 2x } = 3$

$\\ \implies \large \sf \: \frac{ - 2}{ {x}^{2} - 2x } = 3$

$\\ \implies \large \sf \: - 2 = 3( {x}^{2} - 2x)$

$\\ \implies \large \sf \: - 2 = 3 {x}^{2} - 6x$

$\implies \large \sf \: - 2 - 3 {x}^{2} + 6x = 0$

$\\ \implies \large \sf \: - 3 {x}^{2} + 6x - 2 = 0$

$\\ \implies \large \sf \: 3 {x}^{2} - 6x + 2 = 0$

By using formula method :

$\\ \implies \large \sf \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$\\ \implies \large \sf \: x = \frac{6 \pm \sqrt{ {6}^{2} - 4 \times 3 \times 2} }{2 \times 3}$

$\\ \implies \large \sf \: x = \frac{6 \pm \sqrt{36 - 24} }{6}$

$\\ \implies \large \sf \: x = \: \frac{6 \pm \sqrt{12} }{6}$

$\\ \implies \large \sf \: x = \frac{6 \pm2 \sqrt{3} }{6}$

$\\ \implies \large \sf \: x = \frac{ 2(3 \pm \sqrt{3} )}{2 \times 3}$

$\\ \implies \large \sf \: x = \frac{3 \pm \sqrt{3} }{3}$

$\\ \implies \large \sf \ x \: = \frac{ \sqrt{3} \times \sqrt{3} \pm \sqrt{3} }{ \sqrt{3} \times \sqrt{3} }$

$\\ \implies \large \sf \: x = \frac{ \sqrt{3} ( \sqrt{3} \pm1) }{ \sqrt{3} \times \sqrt{3} }$

$\\ \implies \large \sf \: x = \frac{ \sqrt{3} \pm1 }{ \sqrt{3} }$

$\large \boxed{ \red \bigstar{ \bf{root \: are \: x = \frac{ \sqrt{3} + 1 }{ \sqrt{3} } \: and \: x = \frac{ \sqrt{3} - 1}{ \sqrt{3} } }}}$