Bài bất phương trình vô tỉ
David Lee
2020-02-22T22:36:02-05:00
2020-02-22T22:36:02-05:00
http://luyenthi24h.net/vi/news/tai-lieu/bai-bat-phuong-trinh-vo-ti-52.html
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Thứ bảy - 22/02/2020 22:25
Giải bất phương trình:

Hướng dẫn:
Cách 1: Phân tích được:
![\displaystyle \left( {-{{x}^{2}}+2x+3} \right)\left[ {\left( {x+2} \right)\left( {\dfrac{1}{{2\sqrt{{2x+3}}+x+3}}-\dfrac{2}{{x+1+2\sqrt{{x+1}}}}} \right)+\dfrac{1}{{2\sqrt{{2{{x}^{2}}+5x+3}}+3x+3}}-1} \right]\ge 0 \displaystyle \left( {-{{x}^{2}}+2x+3} \right)\left[ {\left( {x+2} \right)\left( {\dfrac{1}{{2\sqrt{{2x+3}}+x+3}}-\dfrac{2}{{x+1+2\sqrt{{x+1}}}}} \right)+\dfrac{1}{{2\sqrt{{2{{x}^{2}}+5x+3}}+3x+3}}-1} \right]\ge 0](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%28+%7B-%7B%7Bx%7D%5E%7B2%7D%7D%2B2x%2B3%7D+%5Cright%29%5Cleft%5B+%7B%5Cleft%28+%7Bx%2B2%7D+%5Cright%29%5Cleft%28+%7B%5Cdfrac%7B1%7D%7B%7B2%5Csqrt%7B%7B2x%2B3%7D%7D%2Bx%2B3%7D%7D-%5Cdfrac%7B2%7D%7B%7Bx%2B1%2B2%5Csqrt%7B%7Bx%2B1%7D%7D%7D%7D%7D+%5Cright%29%2B%5Cdfrac%7B1%7D%7B%7B2%5Csqrt%7B%7B2%7B%7Bx%7D%5E%7B2%7D%7D%2B5x%2B3%7D%7D%2B3x%2B3%7D%7D-1%7D+%5Cright%5D%5Cge+0&bg=ffffff&fg=555555&s=0)
Với chú ý:

Cách 2:
Biến đổi:

Cách 3: Đặt:
. Ta có:
![\displaystyle \left[ {a-\left( {b+1} \right)} \right]\left( {{{a}^{2}}-ab-2{{b}^{2}}} \right)\ge 0 \displaystyle \left[ {a-\left( {b+1} \right)} \right]\left( {{{a}^{2}}-ab-2{{b}^{2}}} \right)\ge 0](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cleft%5B+%7Ba-%5Cleft%28+%7Bb%2B1%7D+%5Cright%29%7D+%5Cright%5D%5Cleft%28+%7B%7B%7Ba%7D%5E%7B2%7D%7D-ab-2%7B%7Bb%7D%5E%7B2%7D%7D%7D+%5Cright%29%5Cge+0&bg=ffffff&fg=555555&s=0)
Tác giả bài viết: David Lee