What is $2 \sqrt{3} - 3 \sqrt{45} - \sqrt{27}$ $2sqrt3 - 3sqrt45 - sqrt27$ ?

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Answer 1 · 2016-02-15T21:18:30

$= - 9 \sqrt{5} - \sqrt{3}$ $=-9sqrt5 -sqrt3$ $or$ $" " or " "$ $- 1 (9 \sqrt{5} + \sqrt{3})$ $-1(9sqrt5 +sqrt3)$

Simplify : $2 \sqrt{3} - 3 \sqrt{45} - \sqrt{27}$ $2sqrt3 -3sqrt45 -sqrt27$

Break down the factor of 45 and 27 so that one of that factor is a perfect square.

$2 \sqrt{3} - 3 \sqrt{9 \cdot 5} - \sqrt{9 \cdot 3}$ $2sqrt3 - 3sqrt(9*5) -sqrt(9*3)$

Square root of 9 is 3 ....

$= 2 \sqrt{3} - 9 \sqrt{5} - 3 \sqrt{3}$ $=2sqrt3 -9sqrt5 -3sqrt3$

Simply the answer ....

$= - 9 \sqrt{5} - \sqrt{3}$ $=-9sqrt5 -sqrt3$ $or$ $" " or " "$

$- 1 (9 \sqrt{5} + \sqrt{3})$ $-1(9sqrt5 +sqrt3)$

What is 2√3−3√45−√272sqrt3 - 3sqrt45 - sqrt27?