\(or\)
\(pt\Leftrightarrow\sqrt[3]{x}-1+\sqrt{x+3}-2=0\)
\(\Leftrightarrow\frac{x-1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}+\frac{x+3-4}{\sqrt{x+3}+2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}+\frac{1}{\sqrt{x+3}+2}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1.\)
\(+x<\)\(1:VT<\)\(\sqrt[3]{1}+\sqrt{1+3}=3=VP\) => vô nghiệm.
\(+x>1:VT>\sqrt[3]{1}+\sqrt{1+3}=3=VP\) => vô nghiệm
\(+x=1:VT=3=VP\) (thoả mãn)
Vậy x=1.