SuperMaths

hình

a,b,c>0

a+b+c$\leq$$\frac{3}{2}$

tim Min P= $\left ( 3+\frac{1}{a} \right )\left ( 3+\frac{1}{b} \right )\left ( 3+\frac{1}{c} \right )$

cho minh sua~ lai cai de ti. 

tim` min P=$\left ( 3+\frac{1}{a}+\frac{1}{b} \right )\left ( 3+\frac{1}{b}+\frac{1}{c} \right )\left ( 3+\frac{1}{a}+\frac{1}{c} \right )$

x,y,z  a,b,c ????

nham`

$\left ( x^2-1 \right )\left ( x+3 \right )\left ( x+5 \right )= \left ( x^2+4x-5 \right )\left ( x^2+4x+3 \right )$

$\Rightarrow \left ( x+2 \right )^{4}-10\left ( x+2 \right )^{2}+9-m=0$

$\left ( x+2 \right )^2= y$ => $y^2-10y+9-m= 0$ $\left\{\begin{matrix} \Delta > 0\Rightarrow 16 < m & & \\ S> 0 & & \\ P> 0 \Rightarrow 9< m& & \end{matrix}\right.$

$\Rightarrow y_1,y_2$ là 2 nghiệm của pt

$\Rightarrow x1= \sqrt{y1}-2, x2= -\sqrt{y1}-2,x3= \sqrt{y2}-2,x4= -\sqrt{y2}-2$

$\Rightarrow \frac{1}{\sqrt{y1}-2}-\frac{1}{\sqrt{y1}+2}+\frac{1}{\sqrt{y2}-2}-\frac{1}{\sqrt{y2}+2}= -1 \Rightarrow \frac{-4}{y1-4}+\frac{-4}{y2-4}= 1\Rightarrow 16= y1y2$

theo viét: $y_1y_2= 9-m$

$\Rightarrow m=-7$

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