Question

[Lizy]

cho `x^2 -2(m+1)x+2m=0`

`x_1 ;x_2` là 2 nghiệm của pt. tìm min \(P=x_1^2+2\left(m+1\right)x_2+4x_1x_2\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\text{Δ}=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot2m\)

\(=4\left(m^2+2m+1\right)-8m\)

\(=4m^2+4>=4>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-2\left(m+1\right)\right]}{1}=2\left(m+1\right)\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{2m}{1}=2m\end{matrix}\right.\)

\(P=x_1^2+x_2\cdot2\left(m+1\right)+4x_1x_2\)

\(=x_1^2+x_2\cdot\left(x_1+x_2\right)+4\cdot2m\)

\(=x_1^2+x_2^2+x_1x_2+8m\)

\(=\left(x_1+x_2\right)^2-x_1x_2+8m\)

\(=\left(2m+2\right)^2-2m+8m\)

\(=4m^2+8m+4+6m\)

\(=4m^2+14m+4\)

\(=4\left(m^2+\dfrac{7}{2}m+1\right)\)

\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{33}{16}\right)\)

\(=4\left(m+\dfrac{7}{4}\right)^2-\dfrac{33}{4}>=-\dfrac{33}{4}\forall m\)

Dấu '=' xảy ra khi \(m+\dfrac{7}{4}=0\)

=>\(m=-\dfrac{7}{4}\)

Answer 2

[Nguyễn Việt Lâm]

\(\Delta'=\left(m+1\right)^2-2m=m^2+1>0;\forall m\)

Phương trình luôn có 2 nghiệm pb với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m\end{matrix}\right.\)

\(P=x_1^2+2\left(m+1\right)x_2+4x_1x_2\)

\(=x_1\left(x_1+x_2\right)-x_1x_2+2\left(m+1\right)x_2+4x_1x_2\)

\(=2\left(m+1\right)x_1+2\left(m+1\right)x_2+3x_1x_2\)

\(=2\left(m+1\right)\left(x_1+x_2\right)+3x_1x_2\)

\(=4\left(m+1\right)^2+6m\)

\(=4m^2+14m+4\)

\(=4\left(m+\dfrac{7}{4}\right)^2-\dfrac{33}{4}\ge-\dfrac{33}{4}\)

\(P_{min}=-\dfrac{33}{4}\) khi \(m=-\dfrac{7}{4}\)