Tóm tắt lại 1 chút:
x²-(2m+5)x+2m+1=0
Δ=b²-4ac=4m²+12m+21
⇒$x_{1}$ = $\frac{-b+\sqrt[]{Δ}}{2a}$ = $\frac{2m+5+\sqrt[]{4m^2+12m+21} }{2}$
⇒$x_{2}$ = $\frac{-b-\sqrt[]{Δ}}{2a}$ = $\frac{2m+5-\sqrt[]{4m^2+12m+21} }{2}$
Ta có: M=|$\sqrt[]{x_1}$ -$\sqrt[]{x_2}$ | = $\sqrt[]{(\sqrt[]{x_1} -\sqrt[]{x_2})^2}$
⇒ M² = ($\sqrt[]{x_1}$ -$\sqrt[]{x_2}$)² = $x_{1}$ +$x_{2}$ - 2$\sqrt[]{x_{1}x_{2}}$
hay M² = $\frac{2m+5+\sqrt[]{4m^2+12m+21} }{2}$+$\frac{2m+5-\sqrt[]{4m^2+12m+21} }{2}$-2.$\sqrt[]{\frac{2m+5+\sqrt[]{4m^2+12m+21} }{2}.\frac{2m+5-\sqrt[]{4m^2+12m+21} }{2}}$
M² = 2m+5 - $\sqrt[]{8m+4}$
M² = 2m+5 - 2.$\sqrt[]{2m+1}$
M² = 2m+1 - 2.$\sqrt[]{2m+1}$ +1 +3
M² = ($\sqrt[]{2m+1}$ -1)²+3
Do ($\sqrt[]{2m+1}$ -1)² ≥ 0 ∀ m
nên ($\sqrt[]{2m+1}$ -1)²+3 ≥ 3 ∀ m
hay M² ≥ 3 ∀ m
⇒ M ≥ √3 ∀ m
Dấu "=" xảy ra ⇔ ($\sqrt[]{2m+1}$ -1)²=0
⇔ $\sqrt[]{2m+1}$ -1=0
⇔$\sqrt[]{2m+1}$=1
⇔2m+1=1
⇔ m=0
Vậy GTNN của M=√3 ⇔ m=0
