gpt

$\frac{1}{\sqrt{x+1}+\sqrt{x}}+\frac{1}{\sqrt{x+2}\sqrt{x+1}}+....+\frac{1}{\sqrt{x+8}+\sqrt{x+7}}=3-x^{2}$
ĐK $: x \ge 0$. PT
$\Leftrightarrow \sqrt{x+1}-\sqrt{x}+\sqrt{x+2}-\sqrt{x+1}+....+\sqrt{x+8}-\sqrt{x+7}=3-x^{2}$
$\Leftrightarrow\sqrt{x+8}-\sqrt{x}+x^{2}-3=0$
$\Leftrightarrow\sqrt{x+8}-3-\sqrt{x}+1+x^{2}-1=0$
$\Leftrightarrow \left ( x-1 \right )\left (\frac{1}{\sqrt{x+8}+3}-\frac{1}{\sqrt{x}+1}+x+1 \right )=0$
Do $x \ge 0 \Rightarrow 1 \ge \frac{1}{\sqrt{x}+1}\Rightarrow \frac{1}{\sqrt{x+8}+3}-\frac{1}{\sqrt{x}+1}+x+1>0$.
Vậy $x=1.$
còn cách nào khác ko b –  rainyday6789 18-09-13 07:39 PM
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. Thanks! –  Trần Nhật Tân 18-09-13 04:51 PM

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