Toán [Toán 9] đại số

L

leminhnghia1

Giải:

Xét $f(x)>0$ với mọi $x$ nên $f(x)$ vô nghiệm $\rightarrow \Delta<0 \rightarrow b^2<4ac$

Ta có:

$\dfrac{3350a+1340c+4ac+4b+1}{b}$

$> \dfrac{3350a+1340c+b^2+4b+1}{b}$

$=\dfrac{3350a}{b}+\dfrac{1340c}{b}+b+\dfrac{1}{b}+4$

\geq $2\sqrt{\dfrac{3350a.1340c}{b^2}}+2+4$

$= 2\sqrt{\dfrac{4ac.1122250}{b^2}}+6$

$> 2\sqrt{1122250}+6$ (Vì $\dfrac{4ac}{b^2}>1$)

$>2016$ (Đpcm)
 
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