Giúp em bài lượng -11

N

nguyenbahiep1

[TEX]sin^6x + cos^6x = 1 - \frac{3}{4}.sin^22x = 1 - \frac{3(1-cos4x)}{8} \\ sin^4x + cos^4x = 1 - \frac{1}{2}sin^22x = 1- \frac{1-cos4x}{4} \\ y = \frac{2 - \frac{3(1-cos4x)}{8}}{2-\frac{1-cos4x}{4}} \\ y = \frac{13 + 3cos4x}{14+2cos4x} = \frac{3}{2} - \frac{8}{2cos4x+14} \\ -1 \leq cos4x \leq 1 \\ \frac{5}{6} \leq y \leq 1 \\ Min y =\frac{5}{6} , cos4x = -1 \\ Max y = 1 , cos4x = 1 [/TEX]
 
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