1 ) \(|\) x+2 \(|\) - \(|\) x-1 \(|\) < x - 3/2
TH1 : x < -2
bpt <=> -x - 2 - ( -x + 1) < x - 3/2
<=> x > -3/2 ( k tm )
TH 2 : -2 \(\le\) x < 1
bpt <=> x + 2 - ( -x+1) < x - 3/2
<=> x < -5/2 (k tm )
TH3 : x \(\ge\) 1
bpt <=> x + 2 - ( x - 1 ) < x - 3/2
<=> x > 9/2 tm
Vậy x > 9/2 .
2 ) x(x - 1)2 \(\ge\) 4 - x
<=> x( x2 - 2x +1 + 1 ) \(\ge\) 4
<=> x3 - 2x2 + 2x - 4 \(\ge\) 0
<=> x2 (x - 2) + 2(x - 2) \(\ge\) 0
<=> (x2 + 2)(x - 2) \(\ge\) 0
Có : x2 + 2 > 0 , với mọi x
=> x - 2 \(\ge\) 0 <=> x \(\ge\) 2 .
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