Question

16 Calculați: (2√5-√2)+(2+√10) +(3√3-5)(3√3+5); 6 (2√2-3√3)+2(√3+3√2)-12 (5√2+1)(5√2-1); (3√2-1)+(3-√7)-(5-3√7)(2√7+3)-4(20-3√7); (√2+ 3√3)² + (3√2-√3)² – (√5 − 2)(√5 +2); 2√5 [(√5-2) + 3(√5-1)]-6(√5 + 1); f (4√3-1)[(√3-3)² + 3(√3-4)] +(3-√3)²; g 3√5-10√5-3√5-[(2-√5)-3(3-√5)]}.​

Answer 1

Răspuns:

$(2\sqrt{5} -\sqrt{2} )+(2+\sqrt{10} )+(3\sqrt{3} -5)(3\sqrt{3} +5)=\\=2\sqrt{5} -\sqrt{2} +2+\sqrt{10} +27-25=\\2\sqrt{5} -\sqrt{2} +4+\sqrt{10}$

$6(2\sqrt{2} -3\sqrt{3} )+2(\sqrt{3} +3\sqrt{2} )-12(5\sqrt{2} +1)(5\sqrt{2} -1)=\\=12\sqrt{2} -18\sqrt{3} +2\sqrt{3} +6\sqrt{2} -12*(25*2-1)=\\=18\sqrt{2} -16\sqrt{3} -12*49=18\sqrt{2} -16\sqrt{3} -588$

$(3\sqrt{2} -1)+(3-\sqrt{7} )-(5-3\sqrt{7} )(2\sqrt{7} +3)-4(20-3\sqrt{7})=\\ =3\sqrt{2} -1+3-\sqrt{7} -(10\sqrt{7} +15-42-9\sqrt{7} )-80+12\sqrt{7}=\\ =3\sqrt{2} -1+3-\sqrt{7} -\sqrt{7} +27-80+12\sqrt{7} =\\=3\sqrt{2} -51+10\sqrt{7}$

$(\sqrt{2} +3\sqrt{3} )^{2} +(3\sqrt{2} -\sqrt{3})^{2} -(\sqrt{5} -2)(\sqrt{5} +2)=\\=2+6\sqrt{6} +27+18-6\sqrt{6} +3-1=\\=2+27+18+3-1=49$

$2\sqrt{5} [(\sqrt{5} -2)+3(\sqrt{5} -1)]-6(\sqrt{5}+1)=\\ =2\sqrt{5} (\sqrt{5} -2+3\sqrt{5} -3)-6\sqrt{5} -6=\\=2\sqrt{5} (4\sqrt{5} -5)-6\sqrt{5} -6=\\=40-10\sqrt{5} -6\sqrt{5} -6=\\=34-16\sqrt{5}$

$(4\sqrt{3} -1)*[(\sqrt{3} -3)^{2} +3(\sqrt{3} -4)]+(3-\sqrt{3})^{2} =\\=(4\sqrt{3}-1)( 3-6\sqrt{3} +9+3\sqrt{3} -12)+9-6\sqrt{3} +3=\\=(4\sqrt{3}-1)*(0-3\sqrt{3} )+9-6\sqrt{3} +3=\\=(4\sqrt{3}-1)*(-3\sqrt{3})+9-6\sqrt{3} +3=\\=-36+3\sqrt{3} +9-6\sqrt{3} +3=\\=-24-3\sqrt{3}$

la ultima nu ai pus inceputul acoladei si nu pot rezolva