Giải bpt
 1,$\sqrt{2x+1}\leq   \frac{2(x+1)}{2-x}$
 2,$\sqrt{x}+\sqrt{x-5}\leqslant \sqrt{5}$
Bài 2 e hoàn toàn làm theo dạng $\sqrt a +\sqrt b \le \sqrt c$ nhưng bài nè đặc biệt nên có thể làm như sau

ĐK $x\ge 5$

BPT $\Leftrightarrow (\sqrt x -\sqrt 5) +\sqrt{x-5}\le 0$

$\Leftrightarrow \dfrac{x-5}{\sqrt x +\sqrt 5} +\sqrt{x-5}\le 0$

$\Leftrightarrow \sqrt{x-5} .\bigg (\dfrac{\sqrt{x-5}}{\sqrt x +\sqrt 5} +1\bigg )\le 0$

$\Leftrightarrow x=5$     vì $\dfrac{\sqrt{x-5}}{\sqrt x +\sqrt 5} +1 >0   \forall x \ge 5$
1)
$\Leftrightarrow \left\{ \begin{array}{l} 2x+1\geq 0\\ \frac{x+1}{2-x} \geq 0\\2x+1\leq \frac{4(x^2+2x+1)}{4-4x+x^2}\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x\geq -\dfrac{1}{2}\\ -1\leq x< 2\\2x^3-11x^2-4x\leq 0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} -\dfrac{1}{2}\leq x<2 \\ x(2x^2-11x-4)\leq 0\end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l} -\dfrac{1}{2}\leq x< 2 \\ \left [ \begin{matrix} 0\le x<2 \\ 2<x \le \dfrac{1}{4}(11+3\sqrt{17}) \\ x\le \dfrac{1}{4}(11-3\sqrt{17}) \end{matrix} \right. \end{array} \right.$

$\Leftrightarrow  \left [ \begin{matrix} 0\le x<2 \\ -\dfrac{1}{2} \le x\le \dfrac{1}{4}(11-3\sqrt{17}) \end{matrix} \right. $
thank a c nhiều :x –  Dân Nguyễn 05-02-14 10:56 PM
tệ quá T_T. phải tranh thủ xem lại bpt mới được:D –  Gió! 05-02-14 03:15 PM
e sai nhiều quá :) để a chữa nhé –  Dép Lê Con Nhà Quê 05-02-14 12:39 PM

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