Toan 8

E

eye_smile

Có:
$\dfrac{4x+3}{{x^2}+1}=\dfrac{{(x+2)^2}}{{x^2}+1}-1$ \geq -1
Dấu "=" xảy ra \Leftrightarrow $x=-2$
 
E

eye_smile

Có:
$\dfrac{4x+3}{{x^2}+1}=4-\dfrac{{(2x-1)^2}}{{x^2}+1}$ \leq 4
Dấu"=" xảy ra \Leftrightarrow $x=\dfrac{1}{2}$
 
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