\(\Leftrightarrow\sqrt[4]{3x+1}-\sqrt[4]{x+1}+\sqrt[4]{4x+1}-\sqrt[4]{x+1}=0\)
\(\Leftrightarrow\frac{3x+1-\left(x+1\right)}{A^3+A^2B+AB^2+B^3}+\frac{4x+1-\left(x+1\right)}{C^3+C^2B+CB^2+B^3}=0\)
\(\left(A=\sqrt[4]{3x+1};\text{ }B=\sqrt[4]{x+1};\text{ }C=\sqrt[4]{4x+1};\text{ }A,B>0;\text{ }C\ge0\right)\)
\(\Leftrightarrow x\left(\frac{2}{....}+\frac{3}{.....}\right)=0\)
\(\Leftrightarrow x=0.\)
ĐK: \(x\ge-\frac{1}{3}\). Ta thấy: (4x + 1) - (3x + 1) = x
\(\sqrt[4]{3x+1}=a\ge0;\sqrt[4]{4x+1}=b>0.\Rightarrow2\sqrt[4]{x+1}=2\sqrt[4]{b^4-a^4+1}\)
PT <=> \(a+b=2\sqrt[4]{b^4-a^4+1}\Leftrightarrow a^4+4a^3b+6a^2b^2+4ab^3+b^4=16b^4-16a^4+16\Leftrightarrow17a^4-15b^4+4a^3b+4ab^3+6a^2b^2=0\)
