BPT

$\sqrt{x + 1} + 1 \geqslant  4x^{2} + \sqrt{3x}$
Phương pháp đoán nghiệm –  ♂Vitamin_Tờ♫ 22-11-13 08:35 PM
ừ.:P. lười nên k nghĩ ra. :P –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 22-11-13 08:32 PM
giải bài này à bạn? –  ♂Vitamin_Tờ♫ 22-11-13 08:27 PM
Điều kiện $x \ge 0$. BPT
$\Leftrightarrow 4x^2-1+\sqrt{3x}-\sqrt{x+1} \le 0$
$\Leftrightarrow (2x-1)(2x+1)+\frac{2x-1}{\sqrt{3x}+\sqrt{x+1} }\le 0$
$\Leftrightarrow (2x-1)\left ( 2x+1+\frac{1}{\sqrt{3x}+\sqrt{x+1} } \right )\le 0$
Do $x\ge 0 \Rightarrow 2x+1+\frac{1}{\sqrt{3x}+\sqrt{x+1} }>0$ nên BPT
$\Leftrightarrow 2x-1 \le 0 \Leftrightarrow x \le 1/2.$
Hãy ấn chữ V dưới chữ đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Các bài tiếp theo mình sẽ sẵn sàng giúp đỡ bạn. –  Trần Nhật Tân 22-11-13 08:35 PM

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