\(PT\Leftrightarrow\left[\left(x+1\right)^2\right]^2-\left(x^2+2\right)^2=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)^2-\left(x^2+2\right)^2=0\)
\(\Leftrightarrow\left(x^2+2x+1-x^2-2\right)\left(x^2+2x+1+x^2+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x^2+2x+3\right)=0\)
Chứng minh được: \(2x^2+2x+3>0\forall x\)
\(\Rightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
\(\Leftrightarrow\left(x^2+2x+1\right)^2-\left(x^2+2\right)^2=0\)
\(\Leftrightarrow\left(x^2+2x+1-x^2-2\right)\left(x^2+2x+1+x^2+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x^2+2x+3\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay x=1/2
\(\Leftrightarrow\left(x^2+2x+1-x^2-2\right)\left(x^2+2x+1+x^2+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x^2+2x+3\ne0\right)=0\Leftrightarrow x=\dfrac{1}{2}\)
