\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3}\left(x+1\right)^2+2\)
TH1: x<-3/2
Pt sẽ là 8/3(x+1)^2+2=-2x-3+1-2x=-4x-2
=>8/3(x^2+2x+1)+2=-4x-2
=>8/3x^2+16/3x+8/3+2+4x+2=0
=>8/3x^2+28/3x+20/3=0
=>x=-1(loại) hoặc x=-5/2(nhận)
TH2: -3/2<=x<1/2
Pt sẽ là 8/3x^2+16/3x+8/3+2=2x+3+1-2x=4
=>8/3x^2+16/3x+2/3=0
=>\(x=\dfrac{-2+\sqrt{3}}{2}\)
TH3: x>=1/2
Pt sẽ là 8/3x^2+16/3x+8/3+2=2x+3+2x-1=4x+2
=>8/3x^2+16/3x+14/3-4x-2=0
=>8/3x^2+4/3x+8/3=0
=>\(x\in\varnothing\)
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