Question

tot exercitiul il stie cineva?

Answer 1

$\displaystyle a).(4 \sqrt{2} - \sqrt{30} )(4 \sqrt{2} + \sqrt{30} )-3( \sqrt{5} +1)^2= \\ =32-30-3( \sqrt{5} ^2+2 \cdot \sqrt{5} \cdot 1+1^2)=32-30-3(5+2 \sqrt{5} +1)= \\ =32-30-15-6 \sqrt{5} -3=-16-6 \sqrt{5} \\ \\ b).4( \sqrt{3} +1)^2- \sqrt{3} (8- \sqrt{3} )-( \sqrt{21} - \sqrt{2} )( \sqrt{21}+ \sqrt{2} )= \\ =4( \sqrt{3} ^2+2 \cdot \sqrt{3} \cdot 1+1^2)-8 \sqrt{3} +3-21+2= \\ =4(3+2 \sqrt{3} +1)-8 \sqrt{3} +3-21+2= \\ =12+8 \sqrt{3} +4-8 \sqrt{3} +3-21+2=0$

$c).(3 \sqrt{5} -1)^2+ \sqrt{180}-(2 \sqrt{3} - \sqrt{5} )(2 \sqrt{3} + \sqrt{5} )= \\ =(3 \sqrt{5} )^2-2 \cdot 3 \sqrt{5} \cdot 1+1^2+6 \sqrt{5} -12+5= \\ =45-6 \sqrt{5} +1+6 \sqrt{5} -12+5=39 \\ \\ d).( \sqrt{14}+3)^2+( \sqrt{7} -3 \sqrt{2} )^2-( \sqrt{15}- \sqrt{7} )( \sqrt{15} + \sqrt{7} )= \\ =( \sqrt{14} )^2+2 \cdot \sqrt{14} \cdot 3+3^2+ (\sqrt{7} )^2-2 \cdot \sqrt{7} \cdot 3 \sqrt{2}+(3 \sqrt{2} )^2-15+7= \\ =14+6 \sqrt{14} +9+7-6 \sqrt{14} +18-15+7=40$

$e).( \sqrt{6} -2)^2+2(1+ \sqrt{6} )^2+( \sqrt{6}-2)( \sqrt{6} +2)= \\ =( \sqrt{6} )^2-2 \cdot \sqrt{6} \cdot 2+2^2+2[1^2+2 \cdot 1 \cdot \sqrt{6} +( \sqrt{6} )^2] +6-4= \\ =6-4 \sqrt{6} +4+2(1+2 \sqrt{6} +6)+6-4= \\ =6-4 \sqrt{6} +4+2+4 \sqrt{6} +12+6-4=26$

$f).2(3 \sqrt{3} +1)^2+3( \sqrt{3} -2)^2-( 11- \sqrt{15} )(11+ \sqrt{15} )= \\ =2[(3 \sqrt{3} )^2+2 \cdot 3 \sqrt{3} \cdot 1+1^2]+3[( \sqrt{3} )^2-2 \cdot \sqrt{3} \cdot 2+2^2]-121+15= \\ =2(27+6 \sqrt{3} +1)+3(3-4 \sqrt{3} +4)-121+15= \\ =54+12 \sqrt{3} +2+9-12 \sqrt{3} +12-121+15=-29$