có pt mô mà giải

ĐKXĐ \(x\ge0,x\ne1\)

ĐKXĐ: \(x\ne1\)

\(x^2+\frac{x^2}{\left(x-1\right)^2}=\frac{5}{4}\)  

\(\Leftrightarrow\frac{x^2\left(x-1\right)^2+x^2}{\left(x-1\right)^2}=\frac{5}{4}\)

\(\Leftrightarrow\frac{x^4-2x^3+2x^2}{x^2-2x+1}=\frac{5}{4}\)

\(\Rightarrow\left(x^4-2x^3+2x^2\right).4=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow4x^4-8x^3+8x^2-\left(5x^2-10x+5\right)=0\)

\(\Leftrightarrow4x^4-8x^3+3x^2+10x-5=0\)

\(\Leftrightarrow4x^3\left(x+1\right)-12x^2\left(x+1\right)+15x\left(x+1\right)-5\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x^3-12x^2+15x-5\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[2x^2\left(2x-1\right)-5x\left(2x-1\right)+5\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)\left(2x^2-5x+5\right)=0\)

Mà \(2x^2-5x+5=2\left(x-\frac{5}{4}\right)^2+\frac{30}{16}>0\forall x\)

Do đó: \(\orbr{\begin{cases}x+1=0\\2x-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=\frac{1}{2}\end{cases}}\)

Có : \(\hept{\begin{cases}2\left(x+y\right)=z^2\Rightarrow2\left(x+y+z\right)+1=z^2+2z+1=\left(z+1\right)^2\\2\left(y+z\right)=x^2\Rightarrow2\left(y+z+x\right)+1=x^2+2x+1=\left(x+1\right)^2\\2\left(z+x\right)=y^2\Rightarrow2\left(z+x+y\right)+1=y^2+2y+1=\left(y+1\right)^2\end{cases}}\)  mà x,y,z không âm.

\(\Rightarrow x=y=z\) .

Thay vào 3 phương trình trên ta có : \(\orbr{\begin{cases}x=y=z=0\\x=y=z=4\end{cases}}\)

Vậy........