ĐKXĐ:...
\(pt\Leftrightarrow\sqrt{x\left(x-2\right)}+\sqrt{x\left(x-4\right)}=\sqrt{x\left(3x+1\right)}\)
\(\Leftrightarrow\sqrt{x}.\sqrt{x-2}+\sqrt{x}.\sqrt{x-4}=\sqrt{x}.\sqrt{3x+1}\)
Thấy x=0 là n0 của pt
Xét \(x\ne0\)
\(pt\Leftrightarrow\sqrt{x-2}+\sqrt{x-4}=\sqrt{3x+1}\)
\(\Leftrightarrow x-2+x-4+2\sqrt{x^2-6x+8}=3x+1\)
\(\Leftrightarrow2\sqrt{x^2-6x+8}=x+7\)
Để pt có n0=> \(x\ge-7\)
Xét x= -7=> \(2\sqrt{\left(-7\right)^2-6.\left(-7\right)+8}=6\sqrt{11}\ne0\Rightarrow x\ne-7\)
Xét \(x>-7\)
\(\Rightarrow4x^2-24x+32=x+7\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{5}{4}\end{matrix}\right.\left(tm\right)\)
Vậy pt có 3 n0...
