$x^2-3x+4=2\sqrt{x-1}_{}$ ĐK: $x_{}$ $\geq1$
$2\sqrt{x-1}=x^2-3x+4_{}$
$⇔4(x-1)=x^4+9x^2+16-6x^3+8x^2-24x_{}$
$⇔4x-4=x^4+17x^2+16-6x^3-24x_{}$
$⇔4x-4-x^4-17x^2-16+6x^3+24x=0_{}$
$⇔28x-20-x^4-17x^2+6x^3=0_{}$
$⇔-x^4+6x^3-17x^2+28x-20=0_{}$
$⇔-x^4+2x^3+4x^3-8x^2-9x^2+18x+10x-20=0_{}$
$⇔-x^3.(x-2)+4x^2.(x-2)-9x.(x-2)+10(x-2)=0_{}$
$⇔-(x-2)(x^3-4x^2+9x-10)=0_{}$
$⇔-(x-2)(x^3-2x^2-2x^2+4x+5x-10)=0_{}$
$⇔-(x-2). [x^2.(x-2)-2x.(x-2)+5.(x-2) ]=0_{}$
$⇔-(x-2)(x-2)(x^2-2x+5)=0_{}$
$⇔-(x-2)^2.(x^2-2x+5)=0_{}$
$⇔(x-2)^2.(x^2-2x+5)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}(x-2)^2=0\\x^2-2x+5=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=2(TMĐK)\\x∉R\end{array} \right.\)
Vậy phương trình trên có nghiệm $x=2_{}$
