1)
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)
x-√(x²-4)≠0;
x²≠x²-4;∅
x²-4≥0;
x²≥4;
x ∈ (-∞;-2] ∪ [2;∞)
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Verified answer
1)
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)
x-√(x²-4)≠0;
x²≠x²-4;∅
x²-4≥0;
x²≥4;
x ∈ (-∞;-2] ∪ [2;∞)