Giải các phương trình sau bằng phương pháp đặt ẩn số phụ:
a) $\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^{2}+5x+3}-16$
b) $(4x-1)\sqrt{x^{2}+1}=2x^{2}+2x+1$
c) $2(1-x)\sqrt{x^{2}+2x-1}=x^{2}-2x-1$
d) $x^{2}-x-6=2\sqrt{x^{3}+8}$
e) $2(x^{2}+2x+3)=5\sqrt{x^{3}+3x^{2}+3x+2}$
e.
Điều kiện: $x^3+3x^2+3x+2\ge0 \Leftrightarrow x\ge-2$
Đặt: $a=\sqrt{x+2};b=\sqrt{x^2+x+1};a,b\ge0$.
Phương trình đã cho trở thành:
        $2(a^2+b^2)=5ab$
$\Leftrightarrow (2a-b)(a-2b)=0$
$\Leftrightarrow \left[\begin{array}{l}2\sqrt{x+2}=\sqrt{x^2+x+1}\\\sqrt{x+2}=2\sqrt{x^2+x+1}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x^2-3x-3=0\\4x^2+3x+2=0\end{array}\right.$
$\Leftrightarrow x=\dfrac{3\pm\sqrt{37}}{2}$, thỏa mãn.
c.
Điều kiện: $x^2+2x-1\ge0$
Đặt $u=\sqrt{x^2+2x-1}\ge0 \Rightarrow u^2=x^2+2x-1$.
Khi đó phương trình trở thành:
       $2(1-x)u=x^2+2x-1-4x$
$\Leftrightarrow u^2-2(1-x)u-4x=0$
$\Leftrightarrow (u-2)(u+2x)=0$
$\Leftrightarrow \left[\begin{array}{l}u=2\\u=-2x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x^2+2x-1=4\\\left\{\begin{array}{l}-2x\ge0\\x^2+2x-1=4x^2\end{array}\right.\end{array}\right.$
$\Leftrightarrow x^2+2x-5=0$
$\Leftrightarrow x=-1\pm\sqrt6$, thỏa mãn.
b.
Đặt $u=\sqrt{x^2+1}\ge1 \Rightarrow u^2=x^2+1$.
Khi đó phương trình trở thành:
      $(4x-1)u=2(x^2+1)+2x-1$
$\Leftrightarrow 2u^2-(4x-1)u+2x-1=0$
$\Leftrightarrow (2u-1)(u-2x+1)=0$
$\Leftrightarrow u=2x-1$, vì $u\ge1$
$\Leftrightarrow \sqrt{x^2+1}=2x-1$
$\Leftrightarrow \left\{\begin{array}{l}2x-1\ge0\\x^2+1=(2x-1)^2\end{array}\right.$
$\Leftrightarrow x=\dfrac{4}{3}$, thỏa mãn.
a.
Điều kiện: $x\ge-1$
Đặt $\sqrt{2x+3}+\sqrt{x+1}=t$ ta có:
       $t^2=3x+4+2.\sqrt{2x+3}.\sqrt{x+1}$
$\Leftrightarrow t^2-4=3x+2.\sqrt{2x+3}.\sqrt{x+1}$
Phương trình trở thành $t=t^2-4-16$
$\Leftrightarrow  t=5$, vì $t>0$.
Với $t=5$ thì $\sqrt{2x+3}+\sqrt{x+1}=5$
Dẫ thấy VT là hàm đồng biến trên $(-1;+\infty)$ nên phương trình có nghiệm duy nhất $x=3$.

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