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1. $\sqrt{35} \ \textless \ 6 \\ 5 \sqrt{6} \ \textgreater \ 4 \sqrt{7}$
2. $= 0,4 \sqrt{2500} + 13 = 0,4*50 + 13 = 33$
$= \sqrt{6*4} - 4 \sqrt{6} + \sqrt{9*6} = 2 \sqrt{6 } - 4 \sqrt{6} + 3 \sqrt{6} = \sqrt{6}$
3. $= ( 3^{2} - ( \sqrt{7}) ^{2}) = 9 - 7 = 2$
$= 3( \sqrt{2} ) ^{2} + \sqrt{50*2} = 3*2 +10= 16$
$= 6^{2} - 12\sqrt{3} + ( \sqrt{3} ) ^{2} = 36 - 12 \sqrt{3} +3 = 39 - 12 \sqrt{3}$
$= 10 \sqrt{3} - 4 \sqrt{16*3} - \sqrt{25*3} = 10 \sqrt{3} - 16 \sqrt{3} - 5 \sqrt{3}= - 11 \sqrt{3}$
4. $= \frac{(2- \sqrt{c})(2+ \sqrt{c)} } { \sqrt{c} (2+ \sqrt{c)} } = \frac{(2- \sqrt{c)}} { \sqrt{c} } = \frac{ \sqrt{c}(2- \sqrt{c)} } {(\sqrt{c} )^{2} } = \frac{2 \sqrt{c} - c } {c}$
5. $= \frac{4(3- \sqrt{15)} + 4( 3+ \sqrt{15} ) } {3 ^{2} - ( \sqrt{15} ^{2} ) } = \frac{12 - 4 \sqrt{15} + 12 + 4 \sqrt{15} } {3 ^{2} - ( \sqrt{15}) ^{2} } = \frac{24} {9 - 15} = \frac{24} {-6} = -4$