III gimnazjum

Policzmy to razem 3

a) Korzystamy z jedynki trygonometrycznej: sin2α+cos2α=1   sin2α−1+cos2α=sin2α+cos2α​−1=1−1=0

Rozwiązanie 1

a) {−8x+3y=511x−4y=−2​ W=∣[−8,3],[11,−4]∣=(−8)⋅(−4)−3⋅11=32−33=(−1) Wx=∣[5,3],[−2,−4]∣=5⋅(−4)−3⋅(−2)=−20+6=(−14) Wy=∣[−8,11],[5,−2]∣=−8⋅(−2)−11⋅5=16−55=−39

a) Sprawdzamy typ układu równań poprzez metodę wyznacznikową. W=<svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='2.400em' viewBox='0 0 333 2400'><path d='M145 15 v585 v1200 v585 c2.667,10,9.667,15,21,15
c10,0,16.667,-5,20,-15 v-585 v-1200 v-585 c-2.667,-10,-9.667,-15,-21,-15
c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v1200 v585 h43z'/></svg>​[4,8],[1,21​]<svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='2.400em' viewBox='0 0 333 2400'><path d='M145 15 v585 v1200 v585 c2.667,10,9.667,15,21,15
c10,0,16.667,-5,20,-15 v-585 v-1200 v-585 c-2.667,-10,-9.667,-15,-21,-15
c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v1200 v585 h43z'/></svg>​=4⋅21​−8⋅1=2−8=(−6) Wyznacznik W jest różny od zera, dlatego układ ten jest oznaczony. Wx=<svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='3.000em' viewBox='0 0 333 3000'><path d='M145 15 v585 v1800 v585 c2.667,10,9.667,15,21,15
c10,0,16.667,-5,20,-15 v-585 v-1800 v-585 c-2.667,-10,-9.667,-15,-21,-15
c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v1800 v585 h43z'/></svg>​1,823,1​​<svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='3.000em' viewBox='0 0 333 3000'><path d='M145 15 v585 v1800 v585 c2.667,10,9.667,15,21,15
c10,0,16.667,-5,20,-15 v-585 v-1800 v-585 c-2.667,-10,-9.667,-15,-21,-15
c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v1800 v585 h43z'/></svg>​=1⋅21​−8⋅3=21​−24=−2321​ Wy=<svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='3.000em' viewBox='0 0 333 3000'><path d='M145 15 v585 v1800 v585 c2.667,10,9.667,15,21,15
c10,0,16.667,-5,20,-15 v-585 v-1800 v-585 c-2.667,-10,-9.667,-15,-21,-15
c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v1800 v585 h43z'/></svg>​4,11,3​<svg xmlns="http://www.w3.org/2000/svg" width='0.333em' height='3.000em' viewBox='0 0 333 3000'><path d='M145 15 v585 v1800 v585 c2.667,10,9.667,15,21,15
c10,0,16.667,-5,20,-15 v-585 v-1800 v-585 c-2.667,-10,-9.667,-15,-21,-15
c-10,0,-16.667,5,-20,15z M188 15 H145 v585 v1800 v585 h43z'/></svg>​=4⋅3−1⋅1=11

Oznaczmy sobie: x-najkrótszy bok trójkąta y-średni bok trójkąta z-najdłuższy bok trójkąta ⎩<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.316em' style='width:0.8889em' viewBox='0 0 888.89 316' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V316 H384z M384 0 H504 V316 H384z'/></svg>⎨<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.316em' style='width:0.8889em' viewBox='0 0 888.89 316' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V316 H384z M384 0 H504 V316 H384z'/></svg>⎧​x+z=10107​y+z=12x+y=10103​​ Wyprowadźmy z pierwszego niewiadomą x i podstawmy ją do trzeciego równania ⎩<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.316em' style='width:0.8889em' viewBox='0 0 888.89 316' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V316 H384z M384 0 H504 V316 H384z'/></svg>⎨<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.316em' style='width:0.8889em' viewBox='0 0 888.89 316' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V316 H384z M384 0 H504 V316 H384z'/></svg>⎧​x=10107​−zy+z=1210107​−z+y=10103​​