Oblicz: -[1-(x-y)²]-[(y-x)²-1] = ................................. (x-3)(x-3)-(x-1)(x-1)+4(x-2)=.......................... Prosze o pomoc ...

-[1-(x-y)²]-[(y-x)²-1]=[(x-y)²-1]-[(y-x)²-1]=0 (x-3)(x-3)-(x-1)(x-1)+4(x-2)=x²-6x+9-x²+2x-1+4x-8=0

W obu przykładach musimu skorzystać ze wzorów skróconego mnożenia: 1. (x - y)² = x² -2xy + y² 2. (x + y)² = x² + 2xy +y² -[1-(x-y)²]-[(y-x)²-1] = -[1 - ( x² -2xy + y²)] - [( y² - 2xy + x²) - 1] = = - [1 - x²+ 2xy - y²] - [y² - 2xy + x² - 1] = = -1 +x² -2xy + y² - y² + 2xy - x² + 1 = = +x² - x² - 2xy + 2xy + y² - y² -1 + 1 = 0 (x-3)(x-3)-(x-1)(x-1)+4(x-2)= (x - 3)² - (x - 1)² + 4x - 8 = = x² -2x*3 + 3² - ( x² -2x*1 + 1²) + 4x - 8 = = x² -6x + 9 - x² + 2x - 1 + 4x - 8 = x²- x² -6x + 2x + 4x + 9 - 1 - 8 = 0