Dat x2 = t ( t>/ 0)
+ t=0 => B=0
+t >0
\(B=\dfrac{t}{t^3+6t^2+12t+8}=\dfrac{1}{t^2+6t+12+\dfrac{8}{t}}=\dfrac{1}{\left(t-1\right)^2+\left(8t+\dfrac{8}{t}\right)+11}\le\dfrac{1}{0+16+11}=\dfrac{1}{27}.\)Bmax = 1/ 27 khi t=1 hay x = +- 1
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