kristinechibende17
Answered

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rationalize the denominator 2√7+√3/3√7-√3​

Sagot :

Answer:

[tex] \frac{2 \sqrt{7} + \sqrt{3}}{3 \sqrt{7} - \sqrt{3} } = \frac{2 \sqrt{7} + \sqrt{3}}{3 \sqrt{7} - \sqrt{3} } . \frac{3 \sqrt{7} + \sqrt{3} }{3 \sqrt{7} + \sqrt{3} } \\ \\ = \frac{(2 \sqrt{7} + 3)(3 \sqrt{7} + \sqrt{3}) }{(3 \sqrt{7} - 3)(3 \sqrt{7} + \sqrt{3} ) } \\ \\ = \frac{(42 + 2 \sqrt{21} + 9 \sqrt{21} + 3 \sqrt{21} )}{63 - 3} \\ \\ = \frac{42 + 2\sqrt{21} + 9 \sqrt{21} + 3 \sqrt{21} }{60} [/tex]

mhanifa

Answer:

  • (2√7+√3)/(3√7-√3)​ =
  • (2√7+√3)(3√7+√3)​/(3√7-√3)​(3√7+√3)​ =
  • (2*3*√7² + 2√21 + 3√21 + √3²)/(9√7² - √3²) =
  • (42 + 3 + 5√21)/(63 - 3) =
  • (45 + 5√21)/60 =
  • (9 + √21)/12
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