Ta có: \(D=\frac{3x^2+2x+1}{2x^2+1}\)
\(=\frac{3x^2+1,5+2x-0,5}{2x^2+1}=\frac{1,5\left(2x^2+1\right)+2x-0,5}{2x^2+1}\)
\(=1,5+\frac{2x-0,5}{2x^2+1}\)
Đặt \(A=\frac{2x-0,5}{2x^2+1}\)
=>\(A\left(2x^2+1\right)=2x-0,5\)
=>\(2A\cdot x^2+A-2x+0,5=0\)
=>\(2A\cdot x^2-2x+A+0,5=0\) (1)
\(\Delta=\left(-2\right)^2-4\cdot2A\cdot\left(A+0,5\right)=4-8A\left(A+0,5\right)=4-8A^2-4A=-8A^2-4A+4\)
\(=-4\left(2A^2+A-1\right)=-4\left(2A^2+2A-A-1\right)=-4\left(A+1\right)\left(2A-1\right)\)
Để (1) có nghiệm thì Δ>=0
=>-4(A+1)(2A-1)>=0
=>(A+1)(2A-1)<=0
=>\(-1\le A\le\frac12\)
=>\(A\ge-1\)
=>\(\frac{2x-0,5}{2x^2+1}\ge-1\forall x\)
=>\(\frac{2x-0,5}{2x^2+1}+1,5\ge-1+1,5=0,5\forall x\)
Dấu '=' xảy ra khi \(\frac{2x-0,5}{2x^2+1}=-1\)
=>\(2x^2+1=-2x+0,5\)
=>\(2x^2+2x+0,5=0\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>2x=-1
=>\(x=-\frac12\)
