Question

Efectuati:
a)√6•(4/√2-3/√3)+(6/√12+9/√27-16/4√8) •1/2√3-√2+√18

b) 2/√2-1+3/√2+1-7/3-√2

Repede va rog!!! Imi trebuie neapărat!!! Dau coroana!

Answer 1

Rezolvare :a)
$\sqrt{6 } \times (2 \sqrt{2 } - \sqrt{3} ) + ( \frac{6}{2 \sqrt{3} } + \frac{9}{3 \sqrt{3} } - \frac{16}{8 \sqrt{2} } ) \times \frac{1}{2 \sqrt{3} } - \sqrt{2 } + 3 \sqrt{2} = \sqrt{6 } \times (2 \sqrt{2} - \sqrt{3}) +( \frac{6}{2 \sqrt{3} } + \frac{9}{3 \sqrt{3} \frac{}{} } - \frac{16}{8 \sqrt{2} } ) \times \frac{1}{2 \sqrt{3} } - \sqrt{2} + 3 \sqrt{2} = 2 \sqrt{12} - \sqrt{18} + ( \frac{3}{ \sqrt{3} } + \frac{3}{ \sqrt{3} } - \frac{2}{ \sqrt{2} }) \times \frac{1}{2 \sqrt{3} } - \sqrt{2} + 3 \sqrt{2} = 4 \sqrt{3} - 3 \sqrt{2} + ( \frac{6}{ \sqrt{3} } - \sqrt{2} ) \times \frac{1}{2 \sqrt{3} } - \sqrt{2} + 3 \sqrt{2} = 4 \sqrt{3} - 3 \sqrt{2} + (2 \sqrt{3} - \sqrt{2} ) \times \frac{1}{2 \sqrt{3} } - \sqrt{2} + 3 \sqrt{2} = 4 \sqrt{3} - 3 \sqrt{2} + \frac{2 \sqrt{3} - \sqrt{2} }{2 \sqrt{3} } - \sqrt{2 } + 3 \sqrt{2} = 4 \sqrt{3} + \frac{(2 \sqrt{3 \ } - \sqrt{2}) \sqrt{3} }{6} - \sqrt{2} = 4 \sqrt{3} + \frac{6 - \sqrt{6} }{6} - \sqrt{2}$
b)
$\frac{2}{ \sqrt{2} - 1} + \frac{3}{ \sqrt{2} + 1 } - \frac{7}{3 \sqrt{2} } = 2( \sqrt{2} + 1) + 3( \sqrt{2} - 1) - (3 + \sqrt{2} ) = 2 \sqrt{2 } + 2 + 3 \sqrt{2} - 3 - 3 - \sqrt{2} = 4 \sqrt{2} - 4$