Matematyka

Autorzy:Wojciech Babiański, Lech Chańko, Dorota Ponczek

Wydawnictwo:Nowa Era

Rok wydania:2014

Usuń niewymierność z mianownika 4.6 gwiazdek na podstawie 5 opinii
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`a)` 

`1/(sqrt2+sqrt3-1)=1/(sqrt2+sqrt3-1)*(sqrt2+sqrt3+1)/(sqrt2+sqrt3+1)=(sqrt2+sqrt3+1)/((sqrt2+sqrt3)^2-1^2)=(sqrt2+sqrt3+1)/(2+2sqrt2*sqrt3+3-1)=` 

`=(sqrt2+sqrt3+1)/(4+2sqrt6)=(sqrt2+sqrt3+1)/(4+2sqrt6)*(4-2sqrt6)/(4-2sqrt6)=((sqrt2+sqrt3+1)(4-2sqrt6))/(4^2-(2sqrt6)^2)=` 

`=((sqrt2+sqrt3+1)*4+(sqrt2+sqrt3+1)*(-2sqrt6))/(16-4*6)=` 

`=(4sqrt2+4sqrt3+4-2sqrt12-2sqrt18-2sqrt6)/(16-24)=` 

`=(4sqrt2+4sqrt3+4-2*sqrt4*sqrt3-2*sqrt9*sqrt2-2sqrt6)/(-8)=` 

`=(4sqrt2+4sqrt3+4-2*2sqrt3-2*3sqrt2-2sqrt6)/(-8)=` 

`=(4sqrt2+4sqrt3+4-4sqrt3-6sqrt2-2sqrt6)/(-8)=` 

`=(-2sqrt2+4-2sqrt6)/(-8)=(sqrt6+sqrt2-2)/4`   

 

 

 

 

`b)` 

`2/(1-sqrt2+sqrt3)=2/(1-(sqrt2-sqrt3))=2/(1-(sqrt2-sqrt3))*(1+(sqrt2-sqrt3))/(1+(sqrt2-sqrt3))=(2+2sqrt2-2sqrt3)/(1^2-(sqrt2-sqrt3)^2)=` 

`=(2+2sqrt2-2sqrt3)/(1-(2-2sqrt6+3))=(2+2sqrt2-2sqrt3)/(1-5+2sqrt6)=(2+2sqrt2-2sqrt3)/(2sqrt6-4)=(1+sqrt2-sqrt3)/(sqrt6-2)=` 

`=(1+sqrt2-sqrt3)/(sqrt6-2)*(sqrt6+2)/(sqrt6+2)=((1+sqrt2-sqrt3)(sqrt6+2))/(sqrt6^2-2^2)=` 

`=(sqrt6+sqrt12-sqrt18+2+2sqrt2-2sqrt3)/(6-4)=(sqrt6+sqrt4*sqrt3-sqrt9*sqrt2+2+2sqrt2-2sqrt3)/2=` 

`=(sqrt6+2sqrt3-3sqrt2+2+2sqrt2-2sqrt3)/2=(sqrt6-sqrt2+2)/2` 

 

 

 

`c)` 

`1/(sqrt2+sqrt3+sqrt5)=1/((sqrt2+sqrt3)+sqrt5)*((sqrt2+sqrt3)-sqrt5)/((sqrt2+sqrt3)-sqrt5)=(sqrt2+sqrt3-sqrt5)/((sqrt2+sqrt3)^2-sqrt5^2)=` 

`=(sqrt2+sqrt3-sqrt5)/(2+2sqrt6+3-5)=(sqrt2+sqrt3-sqrt5)/(2sqrt6)=(sqrt2+sqrt3-sqrt5)/(2sqrt6)*sqrt6/sqrt6=` 

`=(sqrt12+sqrt18-sqrt30)/(2*6)=(sqrt4*sqrt3+sqrt9*sqrt2-sqrt30)/12=(2sqrt3+3sqrt2-sqrt30)/12`