`x^2+2(m+1)x+m-4=0 (1)`
a, Thay `m=-5` vào `(1)` ta đc:
`x²+2(-5+1)x-5-4=0 `
`<=> x² -8x-9=0`
`<=>`\(\left[ \begin{array}{l}x=9\\x=-1\end{array} \right.\)
b, Ta có: `∆' = b'² -ac = (m+1)²-(m-4)`
`= m²+2m+1-m+4`
`= m²+m+5 >0 ∀m`
Theo Viet ta có:
$\left \{ {{x_1+x_2 = -2(m+1)} \atop {x_1.x_2=m-4}} \right.$
Theo đb ta có:
`\frac{x_1}{x_2} + \frac{x_2}{x_1} =-3`
`<=> \frac{x_1²+x_2²}{x_1.x_2}=-3`
`<=> \frac{(x_1+x_2)²-2x_1.x_2}{x_1.x_2}= -3` `<=> \frac{4(m+1)²-2.(m-4)}{m-4}=-3`
`<=> \frac{4m²+8m+4 -2m+8}{m-4}=-3`
`<=> 4m²+6m+12=-3(m-4)`
`<=> 4m²+6m+12+3m-12=0`
`<=> 4m² +9m=0`
`<=>` \(\left[ \begin{array}{l}m=0\\m=\frac{-9}{4}\end{array} \right.\)
