Matematyka

Autorzy:Wojciech Babiański, Lech Chańko, Dorota Ponczek

Wydawnictwo:Nowa Era

Rok wydania:2014

Oblicz 4.64 gwiazdek na podstawie 11 opinii
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`a)\ (1+sqrt2)^2+(1-sqrt2)^2=(1+2sqrt2+2)+(1-2sqrt2+2)=` 

`\ \ \ =(3+2sqrt2)+(3-2sqrt2)=3+2sqrt2+3-2sqrt2=6`  

 

 

`b)\ (sqrt3-1)^2-(2-sqrt3)^2=(3-2sqrt3+1)-(4-4sqrt3+3)=` 

`\ \ \ =(4-2sqrt3)-(7-4sqrt3)=4-2sqrt3-7+4sqrt3=2sqrt3-3` 

 

`c)\ (2sqrt3-3/2)^2-(2sqrt3+3/2)^2=[(2sqrt3-3/2)-(2sqrt3+3/2)]*[(2sqrt3-3/2)+(2sqrt3+3/2)]=` 

`\ \ \ =[2sqrt3-3/2-2sqrt3-3/2]*[2sqrt3-3/2+2sqrt3+3/2]=-6/2*4sqrt3=-3*4sqrt3=-12sqrt3` 

 

`d)\ (4-sqrt5)(4+sqrt5)-(sqrt5-2)(2+sqrt5)=4^2-sqrt5^2-(sqrt5-2)(sqrt5+2)=` 

`\ \ \ =16-5-(sqrt5^2-2^2)=11-(5-4)=11-1=10` 

 

`e)\ (sqrt6-sqrt5)(sqrt6+sqrt5)+(sqrt6-sqrt5)^2=(sqrt6^2-sqrt5^2)+(6-2sqrt6*sqrt5+5)=` 

`\ \ \ =(6-5)+(11-2sqrt30)=1+11-2sqrt30=12-2sqrt30`  

 

`f)\ (2sqrt5-sqrt10)^2-(2sqrt5+1)(1-2sqrt5)=(4*5-4sqrt5*sqrt10+10)-(1+2sqrt5)(1-2sqrt5)=` 

`\ \ \ =20-4sqrt50+10-(1^2-(2sqrt5)^2)=30-4sqrt50-(1-4*5)=` 

`\ \ \ =30-4*sqrt25*sqrt2-1-20=30-4*5*sqrt2-21=9-20sqrt2`