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shaniehh

  • one year ago

Help Me Simplify the expression

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  1. iwillrektyou
    • one year ago
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    What is the problem?

  2. shaniehh
    • one year ago
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  3. iwillrektyou
    • one year ago
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    o_O that's hard, sorry, I'm not up to that yet. :(

  4. shaniehh
    • one year ago
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    and rewrite it in rational exponent form

  5. shaniehh
    • one year ago
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    okay

  6. SolomonZelman
    • one year ago
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    Okay, you basically need to know 3 rules: \(\Large\color{black}{ \displaystyle x^{\color{red}{\rm a}} \times x^{\color{blue}{\rm b}}=x^{\color{red}{\rm a}+\color{blue}{\rm b}}\\[0.5em] }\) \(\Large\color{black}{ \displaystyle x^{\color{red}{\rm a}} \div x^{\color{blue}{\rm b}}=x^{\color{red}{\rm a}-\color{blue}{\rm b}}\\[0.5em] }\) \(\Large\color{black}{ \displaystyle \sqrt[\color{blue}{\rm b}]{x^{\color{red}{\rm a}}}= x^{\color{red}{\rm a}/\color{blue}{\rm b}} }\)

  7. jdoe0001
    • one year ago
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    hmmm actaully... .I kinda miss a couple of fellows, the 3 and 4 at the bottom... lemme rewrite it a bit in a sec

  8. jdoe0001
    • one year ago
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    hmmm \(\large { \cfrac{x^{\frac{2}{3}}\cdot 24y\cdot \sqrt[4]{y^3}}{3x^2\cdot 4\sqrt[3]{x^2}} \\ \quad \\ a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}}\qquad thus \\ \quad \\ \cfrac{x^{\frac{2}{3}}\cdot 24y\cdot \sqrt[4]{y^3}}{3x^2\cdot 4\sqrt[3]{x^2}}\implies \cfrac{x^{\frac{2}{3}}\cdot 24y\cdot y^{\frac{3}{4}}}{3x^2\cdot 4x^{\frac{2}{3}}}\qquad and\ then \\ \quad \\ a^{-\frac{{\color{blue} n}}{{\color{red} m}}} = \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad % radical denominator \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \qquad thus \\ \quad \\ \cfrac{x^{\frac{2}{3}}\cdot 24y^1\cdot y^{\frac{3}{4}}}{3^1x^2\cdot 4^1x^{\frac{2}{3}}}\implies x^{\frac{2}{3}}\cdot 24y^1\cdot y^{\frac{3}{4}}\cdot 3^{-1}x^{-2}\cdot 4^{-1}x^{-\frac{2}{3}} \\ \quad \\ \implies \cfrac{x^{\frac{2}{3}\cdot}\cdot x^{-2}\cdot x^{-\frac{2}{3}}\cdot 24y^1\cdot y^{\frac{3}{4}\cdot } }{3^1\cdot 4^1} }\)

  9. jdoe0001
    • one year ago
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    anyhow, so use the same-base exponent rule and see what you can cancel out from the fraction

  10. shaniehh
    • one year ago
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    It says rewrite it in rational exponent form @jdoe0001

  11. jdoe0001
    • one year ago
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    well, take a look, is all in rational exponents :)

  12. jdoe0001
    • one year ago
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    you just need to simplify it

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