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attashe74 [19]
3 years ago
11

D%20" id="TexFormula1" title=" \frac{3 {w}^{6} {v}^{5} }{15 {w}^{2} v} " alt=" \frac{3 {w}^{6} {v}^{5} }{15 {w}^{2} v} " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ehidna [41]3 years ago
5 0
Breaking down:
\frac{3}{15}  =  \frac{1}{5}
{w}^{6}  \div  {w}^{2}  =  {w}^{4}
{v}^{5}  \div v =  {v}^{4}

Therefore,
\frac{3 {w}^{6} {v}^{5} }{15 {w}^{2}v } \\  =  \frac{{w}^{4} {v}^{4}  }{5}

Hope this helps. - M
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Answer:

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BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

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Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

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⇒ 28x + 3×\sqrt[3]{15x-1}×\sqrt[3]{13x+1} (4\sqrt[3]{x}) = 64x.        

(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

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(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

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⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
3 years ago
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