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FlyinSolo_424

  • 3 years ago

Simplify

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  1. FlyinSolo_424
    • 3 years ago
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    |dw:1341995906264:dw|

  2. nphuongsun93
    • 3 years ago
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    \[\huge \frac{1}{81} - \frac{1}{x^2}\]factor this using a^2-b^2 = (a+b)(a-b)

  3. mayank_mak
    • 3 years ago
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    use \[a ^{2} - b ^{2} = (a-b)\times (a+b)\]

  4. FlyinSolo_424
    • 3 years ago
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    I dont get your jist

  5. Wired
    • 3 years ago
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    \[\Large \frac{1}{Z}*\frac{1}{Z} = \frac{1}{Z^{2}}\] \[81 = Z^{2}\]

  6. mayank_mak
    • 3 years ago
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    |dw:1341996655362:dw|

  7. FlyinSolo_424
    • 3 years ago
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    Ok Someone is going to have to go step by step Im confused.

  8. Wired
    • 3 years ago
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    \[\Large a^{2} = \frac{1}{81} = \frac{1}{9}*\frac{1}{9}\] \[\Large a = \frac{1}{9}\] \[\Large b^{2} = \frac{1}{x^{2}} = \frac{1}{x}*\frac{1}{x}\] \[\Large b = \frac{1}{x}\] \[\Large a^{2}-b^{2} = (a+b)(a-b)\] \[\Large (\frac{1}{9})^{2}-(\frac{1}{x})^{2} = (\frac{1}{9}+\frac{1}{x})(\frac{1}{9}-\frac{1}{x})\]

  9. Wired
    • 3 years ago
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    \[\LARGE \frac{\frac{1}{9}-\frac{1}{x}}{(\frac{1}{9}+\frac{1}{x})(\frac{1}{9}-\frac{1}{x})} = \frac{1}{(\frac{1}{9}+\frac{1}{x})}\]

  10. UnkleRhaukus
    • 3 years ago
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    \[\Large\frac{\frac{1}{9}-\frac1x}{\frac 1{81}-\frac 1{x^2}}\] multiply the nuemerator and denominator of the big fraction by \(81x^2\) \[\Large=\frac{81x^2\left(\frac{1}{9}-\frac1x\right)}{81x^2\left(\frac 1{81}-\frac 1{x^2}\right)}\]distribute \[\Large=\frac{\left(\frac{81x^2}{9}-\frac{81x^2}x\right)}{\left(\frac {81x^2}{81}-\frac {81x^2}{x^2}\right)}\]simplify\[\large=\frac{{9x^2}-81x}{x^2-81}\] factorise\[=\frac{9x(x-9)}{(x+9)(x-9)}\]cancel\[=\frac{9x\cancel{(x-9)}}{(x+9)\cancel{(x-9)}}\]\[\large=\frac{9x}{x+9}\]

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