√(x+10)≥x-2 x+10≥0 x≥-10
x+10≥x²-4x+4 x²-5x-6≥0 x²-5x-6 x1=6 x2=-1 по т. Виета
(x+1)(x-6)≥0
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------------------ -10---- -1--------------------6-------------------
+ - +
x∈[-10;-1]∪[6;∞)
x²+x+4≥16 x²+x+4>0 ← D=1-4*4<0 <br>x²+x-12≥0 по т. Виета x1=-4 x2=3 (x-3)(x+4)≥0
---------------------- -4--------------------------3------------------
+ - +
x∈(-∞;-4]∪[3;∞)
√x-2+√x+6≥4 x-2≥0 x≥2
x-2+x+6+2√(x-2)(x+6) ≥16 2x-12≥-2√(x-2)(x+6)
4x²-48x+144≥4x²+16x-48 64x≥192 x≥3