Tìm m để phương trình sau có $2$ nghiệm dương $x_1,x_2$:
           $(m+2)x^2-(2m-5)x+m-4=0$ 
Sau đó biểu diễn biểu thức $x_1^6 +x_2^6$  theo $S=x_1+x_2  và  P=x_1.x_2 $
Điều kiện: $m\ne-2$
Phương trình có 2 nghiệm dương $x_1 ,x_2$ khi và chỉ khi:
$\left\{ \begin{array}{l} \Delta>0\\x_1+x_2>0\\x_1x_2>0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} (2m-5)^2-4(m+2)(m-4)>0\\ \frac{2m-5}{m+2}>0\\\frac{m-4}{m+2}>0\end{array} \right.$
                                 $\Leftrightarrow \left\{ \begin{array}{l} 57-12m>0\\ \left[\begin{array}{l} m>4\\m<-2 \end{array} \right.\end{array} \right.$ 
                                 $\Leftrightarrow 4<m<\frac{19}{4}$ 

Ta có:
$x_1^6+x_2^6=(x_1^2+x_2^2)^3-3(x_1^2+x_2^2)x_1^2x_2^2$ 
                  $=(S^2-2P)^3-3(S^2-2P)P^2$ 
giải nốt cho e bài kia đi. e đăng rồi mà a chẳng giải j thế? –  cobedangyeu_pro97 16-10-12 09:37 PM

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