Explicație pas cu pas:
[tex]$\mathbf{1)2 \sqrt{3} + 5 \sqrt{3} = (2 + 5) \sqrt{3} = 7 \sqrt{3} }$[/tex]
[tex]$\mathbf{2)5 \sqrt{3} - 2 \sqrt{3} = (5 - 2) \sqrt{3} = 3 \sqrt{3} }$[/tex]
[tex]$\mathbf{3) \sqrt{2} + \sqrt{2} = 1 \sqrt{2} + 1 \sqrt{2} = (1 + 1) \sqrt{2} = 2 \sqrt{2} }$[/tex]
[tex]$\mathbf{4)2 \sqrt{3} \times 3 \sqrt{6} = inmultesti \: 2 \: cu \: 3 \: si \: \sqrt{3} \: cu \: \sqrt{6} = 6 \sqrt{3 \times 6} = 6 \sqrt{18} }$[/tex]
[tex]$\mathbf{6 \sqrt{18} = 6 \sqrt{3 {}^{2} \times 2 } = 6 \times 3 \sqrt{2} = 18 \sqrt{2} }$[/tex]
[tex]$\mathbf{5)2 \sqrt{3} \div 3 \sqrt{6} = \frac{2 \sqrt{3} }{3} \times \sqrt{6} = \frac{2 \sqrt{18} }{3} = \frac{6 \sqrt{2} }{3} = 2 \sqrt{2} }$[/tex]
[tex]$\mathbf{6)2 \sqrt{3} + 6 \times 7 + 2 \sqrt{3} = 4 \sqrt{3} + 42 = dai \: factor \: comun \: pe \: 2 = 2(2 \sqrt{3} + 21) }$[/tex]
[tex]$\mathbf{7) \frac{1 + \sqrt{3} }{ \sqrt{6} } = \frac{(1 + \sqrt{3}) \sqrt{6} }{ \sqrt{6} \sqrt{6} } = \frac{ \sqrt{6} + \sqrt{18} }{6} = \frac{ \sqrt{6} + 3 \sqrt{2} }{6} }$[/tex]
[tex]$\mathbf{8) \frac{ \sqrt{2} + 5 }{ \sqrt{6} + 5 \sqrt{3} } = \frac{( \sqrt{2} + 5)( \sqrt{6} - 5 \sqrt{3}) }{( \sqrt{6} + 5 \sqrt{3})( \sqrt{6} - 5 \sqrt{3} )} = \frac{ \sqrt{12} - 5 \sqrt{6} + 5 \sqrt{6} - 25 \sqrt{3} }{6 - 25 \times 3} = \frac{2 \sqrt{3} - 25 \sqrt{3} }{6 - 75} = \frac{ - 23 \sqrt{3} }{ - 69} = \frac{ \sqrt{3} }{3} }$[/tex]
[tex]$\mathbf{9) \sqrt{294} = \sqrt{ {7}^{2} \times 6 } = 7 \sqrt{6} }$[/tex]
[tex]$\mathbf{10) \sqrt{50} = \sqrt{ {5}^{2} \times 2 } = 5 \sqrt{2} }$[/tex]
[tex]$\mathbf{11) \frac{ \sqrt{21 + \sqrt{225} } }{2} - \frac{ \sqrt{21 - \sqrt{225} } }{2} = \frac{ \sqrt{21 + 15} }{2} - \frac{ \sqrt{21 - 15} }{2} = \frac{ \sqrt{36} }{2} - \frac{ \sqrt{6} }{2} = \frac{6}{2} - \frac{ \sqrt{6} }{2} = 3 - \frac{ \sqrt{6} }{2} }$[/tex]
[tex]$\mathbf{12)(3 \sqrt{2} - 5 \sqrt{32})( - 4 \sqrt{162)} = (3 \sqrt{2} - 20 \sqrt{2})( - 36 \sqrt{2}) = ( - 17 \sqrt{2} )( - 36 \sqrt{2} ) = 17 \sqrt{2} \times 36 \sqrt{2} = 17 \times 2 \times 36 = 1224}$[/tex]
Bafta!
