Áp dụng hệ thức Vi-ét,ta có :

\(\hept{\begin{cases}x_1+x_2=\frac{m-1}{1}=m-1\\x_1x_2=\frac{2m-6}{1}=2m-6\end{cases}}\)

\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{5}{2}\Leftrightarrow\frac{x_1^2+x_2^2}{x_1x_2}=\frac{5}{2}\)

\(\Leftrightarrow\frac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\frac{5}{2}\)

\(\Leftrightarrow\frac{\left(m-1\right)^2-2\left(2m-6\right)}{2m-6}=\frac{m^2-6m+13}{2m-6}=\frac{5}{2}\)

\(\Leftrightarrow2m^2-12m+26=10m-30\Leftrightarrow2m^2-22m+56=0\)

\(\Leftrightarrow\orbr{\begin{cases}m=4\\m=7\end{cases}}\)

Vây .....

ĐK: \(x-\frac{1}{x}\ge0;1-\frac{1}{x}\ge0\)

\(x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\)(1) 

<=> \(x-\sqrt{1-\frac{1}{x}}=\sqrt{x-\frac{1}{x}}\)

=> \(x^2-2x\sqrt{1-\frac{1}{x}}+1-\frac{1}{x}=x-\frac{1}{x}\)

<=> \(x^2-2\sqrt{x^2-x}+1-x=0\)( vì (1) => x \(\ge\)0 ) 

<=> \(\left(x^2-x\right)-2\sqrt{x^2-x}+1=0\)

<=> \(\left(\sqrt{x^2-x}-1\right)^2=0\)

<=> \(\sqrt{x^2-x}=1\)

<=> \(x^2-x-1=0\)

<=> \(\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\\x=\frac{1-\sqrt{5}}{2}\end{cases}}\)

Thử lại ta có:  \(x=\frac{1+\sqrt{5}}{2}\)  thỏa mãn bài toán 

Vậy: 

ĐK: \(\hept{\begin{cases}x\ge2\\y\ge-\frac{1}{3}\end{cases}}\)

\(\sqrt{x-2}+x^3-6x^2+12x=\sqrt{3y+1}+27y^3+27y^2+9y+9\)

<=> \(\sqrt{x-2}+x^3-6x^2+12x-8=\sqrt{3y+1}+27y^3+27y^2+9y+1\)

<=> \(\sqrt{x-2}+\left(x-2\right)^3=\sqrt{3y+1}+\left(3y+1\right)^3\)

<=> \(\left(\sqrt{x-2}-\sqrt{3y+1}\right)+\left[\left(x-2\right)^3-\left(3y+1\right)^3\right]=0\)

<=> \(\frac{x-3y-3}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-3y-3\right)\left[\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right]=0\)

<=> \(\left(x-3y-3\right)\left(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right)=0\)

<=> \(x-3y-3=0\)

vì \(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2>0\)

<=> x = 3y + 3

Thế vào phương trình trên ta có: 

\(2+2\left(3y+3\right)^2-2y^2+3\left(3y+3\right)y-4\left(3y+3\right)-3y=0\)

<=> \(25y^2+30y+8=0\Leftrightarrow\orbr{\begin{cases}y=-\frac{2}{5}\\y=-\frac{4}{5}\end{cases}}\)không thỏa mãn đk 

Vậy hệ vô nghiệm.