Ta có \(x^2+3x-2\sqrt{x^2+3x+12}+4=0\Rightarrow x^2+3x+4=2\sqrt{x^2+3x+12}\Rightarrow\left(x^2+3x+4\right)^2=4\left(x^2+3x+12\right)\Rightarrow x^4+6x^3+17x^2+24x+16=4x^2+12x+48\Rightarrow x^4+6x^3+13x^2+12x-32=0\Rightarrow\left(x-1\right)\left(x+4\right)\left(x^2+3x+8\right)=0\Rightarrow x\in\left\{1,-4\right\}\)(do \(x^2+3x+8>0\))
Vậy \(x\in\left\{1,-4\right\}\)
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