$x+1+\sqrt{x^{2}-4x+1} = 3\sqrt{x}$
Edited by dtlshb, 25-05-2016 - 18:42.
$x+1+\sqrt{x^{2}-4x+1} = 3\sqrt{x}$
Edited by dtlshb, 25-05-2016 - 18:42.
Đk: $x\geq 0$
Ta có:
$x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\Leftrightarrow \left ( 2x+2-5\sqrt{x} \right )+\left ( 2\sqrt{x^2-4x+1}-\sqrt{x} \right )=0\Leftrightarrow$$\left ( 4x^2-17x+4 \right )\left ( \frac{1}{2x+2+5\sqrt{x}}+\frac{1}{2\sqrt{x^2-4x+1}+\sqrt{x}} \right )=0\Leftrightarrow 4x^2-17x+4=0\Leftrightarrow x=4\vee x=\frac{1}{4}$
Edited by takarin1512, 25-05-2016 - 19:56.
$x+1+\sqrt{x^{2}-4x+1} = 3\sqrt{x}$
Đk: $x \ge 0$
Dễ thấy $x=0$ không là nghiệm của pt.
Chia hai vế cho $\sqrt{x}$ ta có pt:
$\sqrt{x}+\dfrac{1}{\sqrt{x}}+\sqrt{x-4+\dfrac{1}{x}}=3$
Đặt: $\sqrt{x}+\dfrac{1}{\sqrt{x}}=a$, ta có pt:
$a+\sqrt{a^2-6}=3$
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$\Leftrightarrow a=\dfrac{5}{2}$
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