гдз 10 клас математика Істер 2018 вправа 17.19
10 клас ➠ математика ➠ Істер
Вправа 17.19
Умова:
Обчисліть границю:
1) lim (x3-9x)/(x2+2x-3);
х→-3
2) lim (2/(x-2)) - (8/(x2-4).
х→2
Відповідь:
1) lim (x3-9x)/(x2+2x-3) = (0/0) = lim (x(x2-9))/((x+3)(x-1)) =
х→-3 х→-3
= lim (x(x-3)(x+3))/((x+3)(x-1)) = lim (x(x-3))/(x-1) =
х→-3 х→-3
= ((-3•(-3-3))/(-3-1) = -9/2 = -4,5
x2 + 2x - 3 = 0
x1 + x2 = -2 x1 = -3
{ {
x1 • x2 = - 3 x2 = 1
2) lim (2/x-2) - (8/(x2-4) = (∞ - ∞) =
х→2
= lim (2/(x-2) - (8/(x-2)(x+2) = lim (2x+4-8)/((x-2)(x+2)) =
х→2 х→2
= lim (2x-4)/((x-2)(x+2)) = lim (2(x-2))/((x-2)(x+2)) =
х→2 х→2
= lim 2/(x+2) = 2/4 = 1/2.
х→2
Умова:
Обчисліть границю:
1) lim (x3-9x)/(x2+2x-3);
х→-3
2) lim (2/(x-2)) - (8/(x2-4).
х→2
Відповідь:
1) lim (x3-9x)/(x2+2x-3) = (0/0) = lim (x(x2-9))/((x+3)(x-1)) =
х→-3 х→-3
= lim (x(x-3)(x+3))/((x+3)(x-1)) = lim (x(x-3))/(x-1) =
х→-3 х→-3
= ((-3•(-3-3))/(-3-1) = -9/2 = -4,5
x2 + 2x - 3 = 0
x1 + x2 = -2 x1 = -3
{ {
x1 • x2 = - 3 x2 = 1
2) lim (2/x-2) - (8/(x2-4) = (∞ - ∞) =
х→2
= lim (2/(x-2) - (8/(x-2)(x+2) = lim (2x+4-8)/((x-2)(x+2)) =
х→2 х→2
= lim (2x-4)/((x-2)(x+2)) = lim (2(x-2))/((x-2)(x+2)) =
х→2 х→2
= lim 2/(x+2) = 2/4 = 1/2.
х→2