a,           $ 5\sqrt{x^3+1}=2(x^2+2)$

b,           $ (x+3)\sqrt{x^2+1}=x^2+3x+1$

c.            $ x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1$

d,             $\sqrt{\dfrac{6}{2-x}}+ \sqrt{\frac{10}{3-x}}=4$
a. Điều kiện: $x\ge -1$.
Đặt: $u=\sqrt{x+1};v=\sqrt{x^2-x+1}, u,v\ge0$, phương trình trở thành:
      $5uv=2(u^2+v^2)$
$\Leftrightarrow (2u-v)(u-2v)=0$
$\Leftrightarrow \left[\begin{array}{l}2u=v\\u=2v\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}4(x+1)=x^2-x+1\\x+1=4(x^2-x+1)\end{array}\right.$
$\Leftrightarrow x=\dfrac{5\pm\sqrt{37}}{2}$
dung rui.may cau duoi ban lam dk ko? –  vipthatt 26-10-13 07:36 PM

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