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T

thuyenbnhv

$(x+10)^2$ \geq 40x
\Rightarrow $\dfrac{x}{(x+10)^2}$ \leq$\dfrac{1}{40}$
dau bang say ra \Leftrightarrow x=10
Chú ý gõ latex
 
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C

congchuaanhsang

$(x+10)^2$ \geq 40x
\Rightarrow $\dfrac{x}{(x+10)^2} \leq$\dfrac{1}{40}$
dau bang say ra \Leftrightarrow x=10
Chú ý gõ latex
Cách khác:

Đặt $x+10$=y\Rightarrow$x=y-10$

$\dfrac{x}{(x+10)^2}$=$\dfrac{y-10}{y^2}$

=$\dfrac{1}{y}-\dfrac{10}{y^2}$

=$-10(\dfrac{1}{y^2}-\dfrac{1}{y}.\dfrac{1}{10}+\dfrac{1}{400})+\dfrac{1}{40}$

=$-10( \dfrac{1}{y}-\dfrac{1}{20} )^2+\dfrac{1}{40}$\leq$\dfrac{1}{40}$
 
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