Giải phương trình: `x^2+4\sqrt{x-1}=2-x`

\(x^2 + 4\sqrt{x-1}=2-x \quad (*)\quad (x\ge 1)\\\Leftrightarrow 4\sqrt{x-1}=2-x-x^2\\\Leftrightarrow 4\sqrt{x-1}=-(-2+x+x^2)\\\Leftrightarrow 16(x-1)=[-(-2+x+x^2)]^2\\\Leftrightarrow 16(x-1)=(x^2 +x+2)^2\\\Leftrightarrow 16(x-1)= (x^2+2x-x-2)^2\\\Leftrightarrow 16(x-1)=[x(x+2)-(x+2)]^2\\\Leftrightarrow 16(x-1)=(x-1)^2(x+2)^2\\\Leftrightarrow 16(x-1)-(x-1)^2(x+2)^2=0\\\Leftrightarrow (x-1)[16-(x-1)(x+2)^2]=0\\\Leftrightarrow \left[ \begin{array}{l}x-1=0\\16(x-1)(x+2)^2=0\end{array} \right.\) 

 Giải phương trình `16(x-1)(x+2)^2=0`

`<=> 16 -(x^2 +4x + 4)(x-1)=0`

`<=> 16-(x^3 -x^2 + 4x^2 -4x + 4x -4)=0`

`<=> 16-x^3 -3x^2 +4=0`

`<=> -x^3 -3x^2 +20 =0`

`<=> -x^2 -2x^2 + 5x^2 +20=0`

`<=> x^2(x-2)+5(x^2-4)=0`

`<=> -(x-2)[x^2+5(x+2)]=0`

`<=> (x-2)(x^2 +5x +10)=0`

\(\Leftrightarrow  x=2,\quad x\notin \mathbb{R}\)

Vậy phương trình `**` có hai nghiệm `1` và `2`