violympic 8

T

thong7enghiaha



$-16x^2-\dfrac{1}{x^2}+15$

$=-(16x^2+\dfrac{1}{x^2}-15)$

$=-(\dfrac{1}{x^2}-2.4x.\dfrac{1}{x}+16x^2-7)$

$=-[(\dfrac{1}{x}-4x)^2-7]$

$=7-(\dfrac{1}{x}-4x)^2$\leq $7$

Vậy GTLN của bt là $7.$
 
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