Question

найти область определения функции : 1)y=sqrt(4-x^2)/sin2x
2)y=(2x^2+3)/x-sqrt(x^2-4)

Answer 1

1) $y=\frac{\sqrt{4-x^2}}{sin2x};$

4-x²≥0;

x²≤4;

-2≤x≤2;

sin2x≠0;

2x≠πk;

x≠(π/2)k; k ∈ Z;

1,57 рад ≈ π/2, то есть -2 < π/2 < 0; и 0 < π/2 < 2;

x ∈ [-2;-π/2) ∪ (-π/2;0) ∪ (0;π/2) ∪ (π/2;2];

2)$\displaystyle y=\frac{2x^2+3}{x-\sqrt{x^2-4}};$

x-√(x²-4)≠0;

x²≠x²-4;∅

x²-4≥0;

x²≥4;

x ∈ (-∞;-2] ∪ [2;∞)