bình phương 2 vế lên là phá được
\(p=\text{|}x\text{|}+\text{|}y\text{|}\)
\(p^2=\left(\text{|}x\text{|}+\text{|}y\text{|}\right)^2=x^2+y^2+2\text{|}xy\text{|}\)
\(P=\text{|}x+y\text{|}\)
\(P^2=\left(\text{|}x+y\text{|}\right)^2\)
nếu \(|x+y|=1\Leftrightarrow\text{(|x+y|)^2}=1\)
nếu \(|x+y|=-1\Leftrightarrow\text{(|x+y|)^2}=1.\)
vậy \(P^2=\text{(|x+y|)^2}=\left(x+y\right)^2\)
\(\Leftrightarrow|x+y|=\sqrt{\left(x+y\right)^2}\)