\(\sqrt{x}-1=mx\sqrt{x}-2mx+1\)

\(\Leftrightarrow mx\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(mx-1\right)=0\)

\(\Leftrightarrow mx-1=0\) (do \(x\ne4\Rightarrow\sqrt{x}-2\ne0\))

Để có x thỏa mãn bài toán

\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\\dfrac{1}{m}\ne1\\\dfrac{1}{m}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>0\\m\ne1\end{matrix}\right.\)

ĐKXĐ: x>=0

\(\sqrt{x} - 1 = mx\sqrt{x} - 2mx + 1\) (1)

Đặt \(t=\sqrt{x}\left(t\ge0\right)\)

=>\(t^2=x;t^3=x\cdot\sqrt{x}\)

(1)=>\(t - 1 = mt^3 - 2mt^2 + 1\)

=>\(mt^3 - 2mt^2 - t + 2 = 0\)

=>\((t - 2)(mt^2 - 1) = 0\)

TH1: t-2=0

=>t=2

=>x=4(nhận)

TH2: \(mt^2-1=0\)

=>\(m\cdot t^2=1\) (2)

Nếu m<=0 thì (2) vô nghiệm

Nếu m>0 thì (2) tương đương: \(t^2=\frac{1}{m}\)

=>\(t=\frac{1}{\sqrt{m}}\)

=>\(x=\frac{1}{m}\)

Để hệ có nghiệm duy nhất thì \(\dfrac{m}{2m}\ne\dfrac{1}{3}\)

=>\(\dfrac{1}{2}\ne\dfrac{1}{3}\)(luôn đúng)

\(\left\{{}\begin{matrix}mx+y=5\\2mx+3y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2mx+2y=10\\2mx+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=4\\mx+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\mx=5-y=5-\left(-4\right)=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-4\\x=\dfrac{9}{m}\end{matrix}\right.\)

\(\left(2m-1\right)\cdot x+\left(m+1\right)\cdot y=m\)

=>\(\dfrac{9}{m}\left(2m-1\right)+\left(m+1\right)\cdot\left(-4\right)=m\)

=>\(\dfrac{9\left(2m-1\right)}{m}=m+4m+4=5m+4\)

=>m(5m+4)=18m-9

=>\(5m^2-14m+9=0\)

=>(m-1)(5m-9)=0

=>\(\left[{}\begin{matrix}m=1\\m=\dfrac{9}{5}\end{matrix}\right.\)