Question

\(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x+2}\)

Answer 1

Điều kiện : \(x\le\dfrac{-\sqrt{2}}{2}\) hoặc \(x\ge\dfrac{3+\sqrt{17}}{2}\)

<=> \(\sqrt{2x^2-1}-\sqrt{2x^2+2x+3}=\sqrt{x^2-x+2}-\sqrt{x^2-3x-2}\)

<=> \(\dfrac{\left(\sqrt{2x^2-1}-\sqrt{2x^2+2x+3}\right)\left(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}\right)}{\sqrt{2x^2-1}+\sqrt{2x^2+2x+3}}\)

<=> \(\dfrac{\left(\sqrt{x^2-x+2}-\sqrt{x^2-3x-2}\right)\left(\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}\right)}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}\)

<=> \(\dfrac{-\left(2x+4\right)}{\sqrt{2x^2-1}+\sqrt{2x^2+2x+3}}=\dfrac{2x+4}{\sqrt{x^2-x+2}+\sqrt{x^2-3x-2}}\)

<=> x = -2 (t/m)

Vậy pt có duy nhất 1 nghiệm x =-2.