Giải phương trình: $\sqrt[]{x^2-1}$ – $\sqrt[]{3x^2+4x+1}$ = (8 – 2x)$\sqrt[]{x+1}$

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1. `\sqrt(x^2-1)-\sqrt(3x^2+4x+1)=(8-2x)\sqrt(x+1)`

   `⇔\sqrt((x-1)(x+1))-3\sqrt((x+1/3)(x+1))-(8-2x)\sqrt(x+1)=0`

   `⇔\sqrt(x+1)(\sqrt(x-1)-3\sqrt(x+1/3)-8+2x)=0`

   ⇔\(\left[ \begin{array}{l}\sqrt(x+1)=0\\\sqrt(x-1)-3\sqrt(x+1/3)-8+2x=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x+1=0\\(x-5)(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4)=0\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=-1\\x-5=0(1/\sqrt((x-1)+2)+2-1/(\sqrt((x+1/3)+4>0)\end{array} \right.\) 

   ⇔\(\left[ \begin{array}{l}x=-1\\x=5\end{array} \right.\) 

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2. Đáp án: $x = – 1; x = 5$

    

   Giải thích các bước giải:

   ĐKXĐ $: x² – 1 ≥ 0 ⇔ x ≤ – 1; x ≥ 1 (1)$

   $ 3x² + 4x + 1 = (x + 1)(3x + 1) ≥ 0 ⇔ x ≤ – 1 ; x ≥ – \dfrac{1}{3} (2)$

   $ x + 1 ≥ 0 ⇔ x ≥ – 1 (3)$

   Kết hợp $(1); (2); (3)$ ĐKXĐ là $:x = – 1; x ≥ 1$

   $ PT ⇔ \sqrt{(x + 1)(x – 1)} – \sqrt{(x + 1)(3x + 1)} – (8 – 2x)\sqrt{(x + 1)} = 0$

   $ ⇔ \sqrt{x + 1}(\sqrt{x – 1} – \sqrt{3x + 1} + 2x – 8) = 0$

   – TH1 $: \sqrt{x + 1} = 0 ⇔ x + 1 = 0 ⇔ x = – 1 (TM)$

   – TH2 $: \sqrt{x – 1} – \sqrt{3x + 1} + 2x – 8 = 0$

   $ ⇔ (\sqrt{x – 1} – 2) + 2(x – 5) – (\sqrt{3x + 1} – 4) = 0$

   $ ⇔ \dfrac{x – 5}{\sqrt{x – 1} + 2} + 2(x – 5) – \dfrac{3(x – 5)}{\sqrt{3x + 1} + 4} = 0$

   $ ⇔ (x – 5)(\dfrac{1}{\sqrt{x – 1} + 2} + 2 – \dfrac{3}{\sqrt{3x + 1} + 4}) = 0 (*)$

   Vì $\dfrac{3}{\sqrt{3x + 1} + 4} < \dfrac{3}{4} < 2 ⇒  \dfrac{1}{\sqrt{x – 1} + 2} + 2 – \dfrac{3}{\sqrt{3x + 1} + 4}> 0$

   $ (*) ⇔ x – 5 = 0 ⇔ x = 5 (TM)$

    

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