tìm cực trị

C

congchuaanhsang

1, Dễ thấy $x^2-5x+7$>0

\RightarrowA=$\dfrac{x^2}{x^2-5x+7}$\geq0

\Rightarrow$A_{min}$=0\Leftrightarrowx=0

Mặt khác để A max thì x phải khác 0

A=$\dfrac{1}{1-\dfrac{5}{x}+\dfrac{7}{x^2}}$

Xét $\dfrac{7}{x^2}-\dfrac{5}{x}+1$=$7( \dfrac{1}{x}-\dfrac{5}{14} )^2+\dfrac{3}{28}$\geq$\dfrac{3}{28}$

\RightarrowA\leq$\dfrac{1}{\dfrac{3}{28}}$=$\dfrac{28}{3}$

$A_{max}$=$\dfrac{28}{3}$\Leftrightarrowx=$\dfrac{14}{5}$
 
C

congchuaanhsang

2, B=$\dfrac{x^2+2x-1}{2x^2+4x+9}$=$\dfrac{7x^2+14x-7}{7(2x^2+4x+9)}$

=$\dfrac{11(x^2+2x+1)-2(2x^2+4x+9)}{7(2x^2+4x+9)}$

=$\dfrac{11(x+1)^2-2(2x^2+4x+9)}{7(2x^2+4x+9)}$

=$\dfrac{11(x+1)^2}{7(2x^2+4x+9)}$-$\dfrac{2}{7}$\geq$-\dfrac{2}{7}$

$B_{min}$=$-\dfrac{2}{7}$\Leftrightarrowx=-1
 
C

congchuaanhsang

3, Đặt C=$\dfrac{x}{x^2+2}=a$

\Leftrightarrow$x=ax^2+2a$\Leftrightarrow$ax^2-x+2a=0$

$\Delta$=$1-8a^2$\geq0\Leftrightarrow$a$\geq$-\dfrac{1}{\sqrt{8}}$=$-\dfrac{\sqrt{2}}{4}$

$C_{min}$=$-\dfrac{\sqrt{2}}{4}$\Leftrightarrowx=$-\sqrt{2}$

 
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