a) \(\dfrac{\sqrt{x}\left(mx^2-3\left(m+1\right)x+2m+3\right)}{x-2}=0\left(1\right)\)
Thay \(m=2\) vào (1) ta được
\(\)\(\dfrac{\sqrt{x}\left(2x^2-9x+7\right)}{x-2}=0\left(x\ge0;x\ne2\right)\)
\(\Leftrightarrow\sqrt{x}\left(2x^2-9x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\2x^2-9x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{7}{2}\end{matrix}\right.\) (thỏa mãn điều kiện)
Vậy \(x\in\left(0;1;\dfrac{7}{2}\right)\) là nghiệm pt đã cho
b) \(\dfrac{\sqrt{x}\left(mx^2-3\left(m+1\right)x+2m+3\right)}{x-2}=0\left(1\right)\left(x\ge0;x\ne2\right)\)
\(\Leftrightarrow\sqrt{x}\left(mx^2-3\left(m+1\right)x+2m+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\mx^2-3\left(m+1\right)x+2m+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\mx^2-3\left(m+1\right)x+2m+3=0\left(2\right)\end{matrix}\right.\)
Để (1) có 3 nghiệm phân biệt khi (2) có 2 nghiệm phân biệt \(x>0;x\ne2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=9\left(m+1\right)^2-4m\left(2m+3\right)>0\\S=\dfrac{3\left(m+1\right)}{m}>0\\P=\dfrac{2m+3}{m}>0\end{matrix}\right.\) \(\left(m\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}9m^2+18m+9-8m^2-12m>0\\m< -1\cup m>0\\m< -\dfrac{3}{2}\cup m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m+9>0\\m< -1\cup m>0\\m< -\dfrac{2}{3}\cup m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+3\right)^2>0\\m< -1\cup m>0\\m< -\dfrac{2}{3}\cup m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne-3\\m< -1\cup m>0\\m< -\dfrac{2}{3}\cup m>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne-3\\m< -1\end{matrix}\right.\) hay \(m>0\) \(\left(3\right)\)
Để (2) có nghiệm \(x\ne2\) khi và chỉ khi
\(m.2^2-3\left(m+1\right).2+2m+3\ne0\)
\(\Leftrightarrow4m-6m-6+2m+3\ne0\Leftrightarrow-3\ne0,\forall m\in R\)
Vậy \(\left\{{}\begin{matrix}m\ne-3\\m< -1\end{matrix}\right.\) hay \(m>0\) thỏa mãn yêu cầu đề bài
