Odpowiedzi

2010-02-07T14:11:58+01:00
3(2-x)²+4(x-5)²= 3(2²-2*2*x+x²)+4(x²-2*x*5+5²)=3(4-4x+x²)+4(x²-10x+25)=12-12x+3x²+4x²-10x+100=112-22x+7x²

(5x-2)(5x+2)-(3x-1)²+4x= ((5x)²-2²-(3x)²-2*(3x)*1+1²)+4x=25x²-4-(9x²-6x+1)+4x=25x²-4-9x²+6x-1+4x=16x²-5+10x

(3y-2)(5y+1)-(3y-4)²+2(5y+1)²+(y-2)(y+2)=(15y²+3y-10y-2)-((3y)²-2*(3y)*4+4²+2((5y)²+2*(5y)*1+1²)+(y²-2²)=(15y²-7y-2)-(9y²-24y+16)+2(25y²+10y+1)+(y²+4)=15y²-7y-2-9y²+24y-16+50y²20y+2+y²-4=57y²+37y-20

(x+2)³-(x-2)(x²+2x+4)= (x³+6x+8)-(x³+2x²+4x-2x²-4x-8)=x³+6x+8-(x³-8)=x³+6x+8-x³+8=6x+16

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