a) 2−2+32=(2−2+3)(2−2−3)2⋅(2−2−3)=(2−2)2−(3)24−22−23=(4−42+2)−34−22−23=
=6−42−34−22−23=3−424−22−23
Usuwamy drugi raz niewymierność z mianownika:
3−424−22−23=(3−42)(3+42)(4−22−23)(3+42)=32−(42)212+162−62−84−63−86=
=9−3212+102−16−63−86=−23−4+102−63−86=234−102+63+86
b) 34−13=(34−1)⋅((34)2+34+1)3⋅((34)2+34+1)=(34)3−133⋅(316+34+1)=4−13⋅(316+34+1)=
=3131⋅(316+34+1)=38⋅2+34+1=232+34+1
c) 2+331=(2+33)(22−233+(33)2)1⋅(22−233+(33)2)=23+(33)34−233+39=8+34−233+39=114−233+39
d) a2−ab+b21−33+392=a2−ab+b2(1−33+39)⋅a+b(1+33)2⋅(1+33)=a313+b3(33)32+233=1+32+233=4221(1+33)=21+33
Oceń to zadanie:
Średnia:
4.57
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