Matematyka

Autorzy:Wojciech Babiański, Lech Chańko, Dorota Ponczek

Wydawnictwo:Nowa Era

Rok wydania:2015

Oblicz 4.63 gwiazdek na podstawie 8 opinii
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`a)\ (3+sqrt2)^2=3^2+2*3*sqrt2+sqrt2^2=9+6sqrt2+2=11+6sqrt2` 

`b)\ (1-sqrt5)^2=1^2-2*1*sqrt5+sqrt5^2=1-2sqrt5+5=6-2sqrt5` 

`c)\ (4-3sqrt2)^2=4^2-2*4*3sqrt2+(3sqrt2)^2=16-24sqrt2+9*2=` 

`\ \ \ =16-24sqrt2+18=34-24sqrt2` 

`d)\ (2sqrt7+3)^2=(2sqrt7)^2+2*2sqrt7*3+3^2=4*7+12sqrt7+9=` 

`\ \ \ =28+12sqrt7+9=37+12sqrt7` 

`e)\ (2sqrt5+sqrt3)^2=(2sqrt5)^2+2*2sqrt5*sqrt3+sqrt3^2=4*5+4sqrt15+3=` 

`\ \ \ =20+4sqrt15+3=23+4sqrt15` 

`f)\ (sqrt6-2sqrt3)^2=sqrt6^2-2*sqrt6*2sqrt3+(2sqrt3)^2=6-4sqrt18+4*3=` 

`\ \ \ =6-4*sqrt9*sqrt2+12=18-4*3sqrt2=18-12sqrt2` 

`g)\ (sqrt8-sqrt2/2)^2=sqrt8^2-2*sqrt8*sqrt2/2+(sqrt2/2)^2=8-sqrt16+2/4=` 

`\ \ \ =8-4+1/2=4 1/2` 

`h)\ (sqrt2/2+sqrt3)^2=(sqrt2/2)^2+2*sqrt2/2*sqrt3+sqrt3^2=2/4+sqrt6+3=3 1/2+sqrt6`