Rozwiąż nieróności:
1) 1-(3x-2)² ≥ (x-36)÷3 - 9(x+1)(x-1)
Sprowadź wyrażenie : [y³ ÷ (y² · y-³)]⁴ : [ (1/y)⁴ · 1/y-²]-³ ; y≠0 do najprostszej postaci.
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Zad.1
1-(3x-2)²≥(x-36)/3-9(x+1)(x-1)
1-(9x²+4-12x)≥¹/₃x-12-9(x²-1)
-9x²+12x-3≥-9x²+¹/₃x-3
12x≥¹/₃x
11²/₃x≥0
x≥0
x∈<0;+∞)
Zad.2
(y³:(y²·y⁻³))⁴:((1/y)⁴·(1/y)⁻²)⁻³=(y³:y⁻¹)⁴:(y⁻⁴·y²)⁻³=(y⁴)⁴:(y⁻²)⁻³=y¹⁶:y⁶=y¹⁰