a, \(3x^3-5x^2-x-2>0\)
\(< =>3x^3+x^2+x-6x^2-2x-2>0\)
\(< =>x\left(3x^2+x+1\right)-2\left(3x^2+x+1\right)>0\)
\(< =>\left(x-2\right)\left(3x^2+x+1\right)>0\)
có \(3x^2+x+1=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}\right)=3\left[x^2+2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{35}{36}\right]\)
\(=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{35}{36}\right]>0=>x-2>0< =>x>2\)
b, \(A=2x^2+y^2-2xy-2x+3\)
\(A=x^2-2xy+y^2+x^2-2x+1+2\)
\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\)
dấu"=" xảy ra<=>\(x=y=1\)
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