1 câu trả lời
$x^2-x-4=2\sqrt{x-1}\left(1-x\right)$
$⇔x^2-x-4=2\left(1-x\right)x-1$
$⇔\left(x^2-x-4\right)^2=4\left(1-x\right)^2\left(x-1\right)$
$⇔\left(x^2-x-4\right)^2=4x-4-8x^2+8x+4x^3-4x^2$
$⇔\left(x^2-x-4\right)^2=12x-4-12x^2+4x^3$
$⇔\left(x^2-x-4\right)^2-12x+4+12x^2-4x^3=0$
$⇔x^4-6x^3+5x^2-4x+20=0$
$⇔\left(x-2\right)\left(x-5\right)\left(x^2+x+2\right)=0$
\(⇔\left[ \begin{array}{l}x-2=0\\x-5=0(ktm)\quad \\x^2+x+2=0(ktm)\end{array} \right.\)
$⇔x=2$
$\text{Vậy pt có nghiệm S = {2}}$
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