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Scoala nr. 2 Ghirdoveni

Scoala nr. 2 Ghirdoveni. Formule de calcul prescurtat. Formula (a-b)( a+b )=a 2 -b 2 ; a,b Є R ( a+b )(a-b)=a 2 -ab+ab-b 2 =a 2 -b 2. Observatii !!!

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Scoala nr. 2 Ghirdoveni

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  1. Scoala nr. 2 Ghirdoveni

  2. Formule de calcul prescurtat

  3. Formula (a-b)(a+b)=a2-b2 ; a,bЄ R(a+b)(a-b)=a2-ab+ab-b2=a2-b2 Observatii !!! • Asadar , produsuldintresumasidiferenta a doitermeniesteegal cu diferentapatratelorcelordoitermeni. Pentruefectuarearapida a unorasemeneacalcule , retinemaceasta formula (a-b)(a+b)=a2-b2,oricarearfia,bЄR

  4. Formule • (a+b)3=(a+b)(a+b)(a+b)=a3+ab2+ba2+ab2+ba2+ab2+ba2+b3=a3+3ab2+3a2b+b3 • (a-b)3=(a-b)(a-b)(a-b)=a3-3a2b+3ab2-b3 • a3-b3= (a-b)(a2+ab+b2) • a3+b3=(a+b)(a2-ab+b2) • a3+b3+c3=(a+b+c)3-3(a+b+c)(a+b+c)(a+b+c)

  5. Exercitii Propunclaseisprerezolvareurmatoareloraplicatii: a) Sa se aflenumerelenaturale,xsi y pentru care: 4x2+y2-4x+2-2y=0 b)Determinatinaturatriunghiului in care are loc relatia: a2+b2+c2=ab+bc+ac c) Care estevaloarea minima a expresiei E(x) sipentrucevaloare a lui ,,x’’ se obtine: E ( x ) = x 2 ( x 2 - 2 ) - 3

  6. Solutia(a) Expresia se poatescrie : (4x2-4x+1)+(y2-2y+1)=0 sau ( 2 x – 1 ) 2 + ( y – 1 )2=0 De underezulta:x=1/2 si y= 1

  7. Solutia (b) Egalitatea se inmulteste cu 2, se trecetotulintr-un membrusi se formeaza o suma de patrateperfecte: (a-b)2+(b-c)2+(a-c)2=0 de underezulta: a-b=0 b-c=0 a-c=0 Decia=b=c , triunghiulesteechilateral

  8. Solutia (c) Expresia se poatescrie: E(x)=(x4-2x2+1)-4=(x2-1)2-4 Cum un patrat perfect esteintotdeaunamai mare sauegal cu zero,valoarea minima se obtinecandpatratuleste zero. Deci , candx=1saux=-1 Valoarea minima a expresieieste-4

  9. Aplicatii 1.

  10. Exercitii Determinativaloarea minima a expresiei E(x,y) pentruoricenumerereale x si y , undeE(x,y)=√x2-6x+9 + √9y2+6y+10 Solutie : sau Expresia are valoarea minima cand: x=3 si y=-1/3

  11. Aplicatii ale sumelor de patrate in geometrie • Sa se arate ca paralelipipeduldreptunghic in care St=2d2 (St- aria totalaiar d – diagonalaparalelipipedului) estecub. Solutia: Din formula arieitotale a paralelipipeduluisi a diagonaleiacesteiadeducem: 2(ab+bc+ac)=2(a2+b2+c2) rezulta (a-b)2+(b-c)2+(a-c)2=0 Deci,a=b=c

  12. 2.Calculativolumulunuiparalelipipeddreptunghic de dimensiunia,b,c, care verifica relatia:12a+15b+16c=625, si are diagonala 25. Solutia: Fie d- diagonalaparalelipipedului. Din ipotezaavem: 12a+15b+16c=625 sau 12a+30b=32c=2(a2+b2+c2) rezulta a2+b2+c2+a2+b2+c2-24a-30b-32c=0rezulta625+a2+b2+c2-24a-30b-32c=0 Care se poatescrie (a-12)2+(b-15)2+(c-16)2=0 De undeobtinem: a=12, b=15, c=16,si volumul, V=12*15*16=2880 m3

  13. Probleme cu teoremaluiPitagora 1.Fie triunghi ABC dreptunghic in Â: a) Dacalungimilecatetelor AB si AC sunt 4 cm respectic 3 cm determinatilungimeaipotenuzei BC. b) Dacacateta AC=6cm, iaripotenuza BC=10 cm, determinatilungimeacatetei AB.

  14. Solutia (a) AplicamteoremaluiPitagoraastfel: BC2=AC2+AB2 Inlocuim: BC2=42+32 BC2=16+9 BC2=25 cm, rezulta BC=5 cm

  15. Solutia (b) AplicamteoremaluiPitagoraastfe: AB2=BC2-AC2 Inlocuim: AB2=102-62 AB2=100-36 AB2=64 cm, de undeAB=8 cm

  16. Vamultumimpentruvizionare. Sper ca vaplacut. Prezentarea a fostfacuta de douaeleve ale Scolii Nr.2 Ghirdoveni : Angelescu Ana Maria siIonescuIuliana pentru a vaarata cat de distractivasifrumoasapoatefimatematica . Spersanevotati:-* 

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