Chủ đề này có 1 trả lời

[duong7cvl]

GPT:$2x+ (x+1)\sqrt{x^2+2x+3}+(x+2)\sqrt{x^2+4x+6}+3=0$

[leminhnghiatt]

GPT:$2x+ (x+1)\sqrt{x^2+2x+3}+(x+2)\sqrt{x^2+4x+6}+3=0$

 

$\iff 4x+6+2(x+1)\sqrt{x^2+2x+3}+2(x+2)\sqrt{x^2+4x+6}=0$

 

$\iff 10x+15+(x+1)(2\sqrt{x^2+2x+3}-3)+(x+2)(2\sqrt{x^2+4x+6}-3)=0$

 

$\iff 10x+15+\dfrac{(x+1)(2x+1)(2x+3)}{2\sqrt{x^2+2x+3}+3}+\dfrac{(x+2)(2x+5)(2x+3)}{2\sqrt{x^2+4x+6}+3}=0$

 

$\iff (2x+3)(5+\dfrac{2x^2+3x+1}{2\sqrt{x^2+2x+3}+3}+\dfrac{2x^2+9x+10}{2\sqrt{x^2+4x+6}+3})=0$

 

$\iff x=\dfrac{-3}{2}$    v    $5+\dfrac{2x^2+3x+1}{2\sqrt{x^2+2x+3}+3}+\dfrac{2x^2+9x+10}{2\sqrt{x^2+4x+6}+3}=0 \ (*)$

 

Xét $(*)$ ta có:

$\dfrac{2x^2+3x+1}{2\sqrt{x^2+2x+3}+3}+1=\dfrac{2x^2+3x+4+2\sqrt{x^2+2x+3}}{2\sqrt{x^2+2x+3}+3} >0$

 

$\dfrac{2x^2+9x+10}{2\sqrt{x^2+4x+6}+3}+1=\dfrac{2x^2+9x+13+2\sqrt{x^2+4x+6}}{2\sqrt{x^2+4x+6}+3} >0$

 

$3 >0$

 

Cộng các bđt lại ta có: 

 

$5+\dfrac{2x^2+3x+1}{2\sqrt{x^2+2x+3}+3}+\dfrac{2x^2+9x+10}{2\sqrt{x^2+4x+6}+3} >0$ 

 

Vậy (*) vô nghiệm, nghiệm pt là $x=\dfrac{-3}{2}$

Bài viết đã được chỉnh sửa nội dung bởi leminhnghiatt: 10-02-2016 - 12:51