giup minh bai luong giac nay voi

P

passer

[TEX](m+1)tan^4x - 3m(a-tan^2x )tan^2x + \frac{4m}{cos^2x}[/TEX]
[TEX]\Leftrightarrow (2m+1)tan^4x + 5mtan^2x +4m = 0[/TEX]

[TEX]\triangle = 25m^2 - 32m^2 - 16m = -7m^2 -16m[/TEX]

Để pt có no thì [TEX]\triangle \geq 0 \Leftrightarrow m \in [\frac{-16}{7},0][/TEX]

[TEX]tan^2x=\frac{-5m \pm \sqrt{-7m - 16m}}{4m + 2}[/TEX]

theo ycbt [TEX]\Rightarrow tan^2x > 0[/TEX]
[TEX]* -5m + \sqrt{-7m - 16m}>0[/TEX]
[TEX]\Leftrightarrow m \in (\frac{1}{2},0) [/TEX] (loại)

[TEX]* -5m - \sqrt{-7m - 16m}>0 m<0[/TEX]
[TEX]\Leftrightarrow m \in (-\infty , 0)[/TEX]

từ các ĐK [TEX]\Rightarrow m \in [\frac{-16}{7}, 0)[/TEX] thỏa ycbt
 
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