Решить неравенство √(x²-8x+6) + √(4x-x²-3) < x − 1

Question

Решить неравенство √(x²-8x+6) + √(4x-x²-3) < x − 1

Answer 1

ОДЗ
{x-1>0⇒x>1
{x²-8x+6≥0  D=64-24=40 x1=4-√10 U xx2=4+√10⇒x≤4-√10 U x≥4+√10{x²-4x+3≤0  x1+x2=4 U x1*x2=3⇒x1=1 U x2=3⇒1≤x≤ 3

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                                        \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
-------[4-√10]---------------[1]-------[3]--------[4+√10]------------

Answer 2

ОДЗ
{x-1>0⇒x>1
{x²-8x+6≥0  D=64-24=40 x1=4-√10 U xx2=4+√10⇒x≤4-√10 U x≥4+√10
{x²-4x+3≤0  x1+x2=4 U x1*x2=3⇒x1=1 U x2=3⇒1≤x≤ 3
\\\\\\\\\\\\\\\\\\                     \\\\\\\\\\\\           \\\\\\\\\\\\\\\\\\\\\\\\\\\\\
-------[4-√10]---------------[1]-------[3]--------[4+√10]------------
                                        \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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