Cho PT: $x^2-(2m-3)x+m^2-3m=0.$ Xác định $m$ để phương trình có 2 nghiệm $x_1;x_2$ thoả mãn
$1<x_1<x_2<6$
Cách khác $\triangle =9\Rightarrow .....$ pt có 2 nghiệm phân biệt
$\Rightarrow x_2=m;x_1=m-3$
Vì $1<x_1<6 \Rightarrow 4<m<9$
  $1<x_2<6 \Rightarrow 1<m<6$
$\Rightarrow 4 <m<6$
Hehe cách này ngắn hơn cái trên mà sao không làm nhỉ
$\triangle =9>0$ nên pt luôn có hai nghiệm pb
theo viet
$x_1+x_2=2m-3$
 $x_1.x_2=m^2-3m$
 ta có $x_1.x_2>0\Rightarrow m>\frac{3}{2}$ (1)
ta có $2<x_1+x_2<12\Rightarrow 2,5<m<7,5$ (2)
ta xét $(x_1-1)(x_2-1)>0\Rightarrow m< 1;m> 4$ (3)
Xét $(6-x_1)(6-x_2)>0\Rightarrow m>9;m<6$ (4)
Từ (1);(2);(3);(4) $\Rightarrow 4<m<6$


ca còn yêu đời lắm nên chưa chết dc :D –  ๖ۣۜJinღ๖ۣۜKaido 07-05-16 09:08 PM
cách này dễ chết nèlớp muội mấy đứa bị chết đấy –  ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 07-05-16 06:22 PM

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