a) 2a(a-4)-3(a+3)² = 2a²-8a-3(a²+6a+9) = 2a²-8a-3a²-18a-27 = -a²-26a-27 = -(-2)²-26*(-2)-27 = 4 + 52 - 27 = 29 b) (b-1)(1 +b)-2(b +3)(b-5) = (b-1)(b+1)-2(b+3)(b-5) = b²-1²-2(b²-5b+3b-15) = b²-1²-2b²+10b+6b+30 = 3²-1-2*(3²)+10*3+6*3+30 = 9-1-18+30+18+30 = 68 c) 2(c-5)²-(2c-3)(4 +c) = 2(c²-10c+25)-(8c+2c²-12-3c) = 2c²-20c+50-8c-2c²+12+3c = -25c+62 = -25*½+62 = -12,5+62 = 49,5 (osoba wyżej podpunkt c ma tak samo tylko przy wyniku źle napisała, bo -25*½ = -12,5 a nie 12,5 :) )
2a(a-4)-3(a + 3)^2 =2a^2-8a-3(a^2+6a+9)=2a^2-8a-3a^2+18a+27=-a^2+10a+27 dla a=-2 (-2)^2+10(-2)+27=4-20+27=11 (b-1)(1 +b)-2(b +3)(b-5)=b^2-1-2(b^2-5b+3b-15)=b^2-1-2b^2+4b+30=-b^2+4b+30 dla b=3 (-3)^2+4*3+30=9+12+30=51 2(c-5)^2-(2c-3)(4 +c)=2(c^2-10c+25)-(8c+2c^2-12-3c)=2c^2-20c+50-8c-2c^2+12+3c=-25c+62 dla=1/2 -25(1/2)+62=12,5+62=74,5