8 szkoły podstawowej

Matematyka z kluczem 8

Dane są wyrażenia: A=2xy−(x+2y)(x−3y)   B=3xy−(y+3x)(y−2x)   C=(x−1)(2x+3)+3−x   uprośćmy te wyrażenia: A=2xy−(x2−3xy+2xy−6y2)=2xy−x2+3xy−2xy+6y2=−x2+3xy+6y2

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Wiemy, że: -rower jest o 500 zł droższy od hulajnogi -gdyby rower był 20 zł tańszy to hulajnoga kosztowałaby 5 razy tyle x  - cena roweru zatem

Rozwiążmy równanie I: 2x+2​−1=32x+4​  ∣⋅6

Wiemy, że: A=5−2x   B=6x+4   C=1−3x   a) A⋅C=B⋅x   zatem: (5−2x)⋅(1−3x)=(6x+4)⋅x

Wprowadźmy następujące oznaczenia x  -wiek mamy Jurka x+51​x=56​x    - wiek taty Jurka (jest o  51​   starszy od mamy)  Wiemy, że Jurek i jego rodzice mają w sumie 100 lat, zatem:

Wprowadźmy następujące oznaczenia: x  - liczba mrówek w drugim mrowisku 2x  - liczba mrówek w pierwszym mrowisku x+4000  - liczba mrówek w trzecim mrowisku wiemy, że w tych trzech mrowiskach w sumie znajduje się 9000 mrówek zatem:

wprowadźmy następujące oznaczenia: x  - długość tego prostokąta 50%x  - szerokość tego prostokąta Wyznaczmy obwód tego prostokąta: L1​=2x+2⋅50%x=2x+2⋅0,5x=2x+x=3x

x  -   początkowa cena gitary Wiemy, że gitara najpierw podrożała o 15%: x+15%x=x+0,15x=1,15x,     a następnie jej cenę obniżono o 15%: 1,15x−15%⋅1,15x=1,15x−0,15⋅1,15x=1,15x−0,1725x=0,9775x,     i w efekcie tych zmian kosztuje o 13,50 zł mniej niż na początku, a zatem:

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