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FlyinSolo_424
 3 years ago
Simplify
FlyinSolo_424
 3 years ago
Simplify

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FlyinSolo_424
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1341995906264:dw

nphuongsun93
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \frac{1}{81}  \frac{1}{x^2}\]factor this using a^2b^2 = (a+b)(ab)

mayank_mak
 3 years ago
Best ResponseYou've already chosen the best response.1use \[a ^{2}  b ^{2} = (ab)\times (a+b)\]

FlyinSolo_424
 3 years ago
Best ResponseYou've already chosen the best response.0I dont get your jist

Wired
 3 years ago
Best ResponseYou've already chosen the best response.3\[\Large \frac{1}{Z}*\frac{1}{Z} = \frac{1}{Z^{2}}\] \[81 = Z^{2}\]

mayank_mak
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1341996655362:dw

FlyinSolo_424
 3 years ago
Best ResponseYou've already chosen the best response.0Ok Someone is going to have to go step by step Im confused.

Wired
 3 years ago
Best ResponseYou've already chosen the best response.3\[\Large a^{2} = \frac{1}{81} = \frac{1}{9}*\frac{1}{9}\] \[\Large a = \frac{1}{9}\] \[\Large b^{2} = \frac{1}{x^{2}} = \frac{1}{x}*\frac{1}{x}\] \[\Large b = \frac{1}{x}\] \[\Large a^{2}b^{2} = (a+b)(ab)\] \[\Large (\frac{1}{9})^{2}(\frac{1}{x})^{2} = (\frac{1}{9}+\frac{1}{x})(\frac{1}{9}\frac{1}{x})\]

Wired
 3 years ago
Best ResponseYou've already chosen the best response.3\[\LARGE \frac{\frac{1}{9}\frac{1}{x}}{(\frac{1}{9}+\frac{1}{x})(\frac{1}{9}\frac{1}{x})} = \frac{1}{(\frac{1}{9}+\frac{1}{x})}\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[\Large\frac{\frac{1}{9}\frac1x}{\frac 1{81}\frac 1{x^2}}\] 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\[\large=\frac{{9x^2}81x}{x^281}\] factorise\[=\frac{9x(x9)}{(x+9)(x9)}\]cancel\[=\frac{9x\cancel{(x9)}}{(x+9)\cancel{(x9)}}\]\[\large=\frac{9x}{x+9}\]
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