Question

[Triều Nguyễn Quốc]

Tìm x : \(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3\left(x+1\right)^2+2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\Leftrightarrow\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3x^2+6x+5}\)

TH1: x<-3/2

Pt sẽ là \(\dfrac{8}{3x^2+6x+5}=-2x-3+1-2x=-4x-2\)

=>(3x^2+6x+5)(-4x-2)=8

=>-12x^3-6x^2-24x^2-12x-20x-10-8=0

=>-12x^3-30x^2-32x-18=0

=>x=-1,35(loại)

TH2: -3/2<=x<1/2

Pt sẽ là \(\dfrac{8}{3x^2+6x+5}=2x+3+1-2x=4\)

=>3x^2+6x+5=2

=>3x^2+6x+3=0

=>x=-1(nhận)

TH3: x>=1/2

=>\(\dfrac{8}{3x^2+6x+5}=2x+3+2x-1=4x+2\)

=>(3x^2+6x+5)(4x+2)=8

=>12x^3+6x^2+24x^2+12x+20x+10-8=0

=>12x^3+30x^2+32x+2=0

=>x=-0,06(loại)