Temat:
MatematykaAutor:
spirit22Utworzono:
1 rok temua)
[tex]\left \{ {{y=x^2-3} \atop {y=-x^2+4x-3}} \right. \\x^2-3=-x^2+4x-3\\x^2+x^2-4x-3+3=0\\2x^2-4x=0\\\Delta=(-4)^2-4*2*0=16\\\sqrt{\Delta}=4\\x_1=\frac{4-4}{4}=0\\x_2=\frac{4+4}4=\frac{8}4=2\\y_1=0^2-3=-3\\y_2=2^2-3=4-3=1\\\\\text{Rozwiazaniem ukladu rownan sa punkty: }\\(0, -3), (2, 1)[/tex]
b)
[tex]\left \{ {{y=-\frac12x^2+2} \atop {y=\frac12x^2-2x+2} \right. \\-\frac12x^2+2=\frac12x^2-2x+2\\-\frac12x^2-\frac12x^2+2x+2-2=0\\-x^2+2x=0\\\Delta=2^2-4*(-1)*0=4\\\sqrt{\Delta}=2\\x_1=\frac{-2-2}{-2}=\frac{-4}{-2}=2\\x_2=\frac{-2+2}{-2}=0\\\\y_1=-\frac12*2^2+2=-\frac12*4+2=-2+2=0\\y_2=-\frac12*0^2+2=2\\\\\text{Rozwiazaniem ukladu rownan sa punkty: }\\(2, 0), (0, 2)[/tex]
Autor:
jeramiahupb2
Oceń odpowiedź:
0