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Theo định lý Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2+x_3=m+2\\x_1x_2+x_1x_3+x_2x_3=3m\\x_1x_2x_3=1\end{matrix}\right.\)
\(P=x_1^2+x_2^2+x_3^2=\left(x_1+x_2+x_3\right)^2-2\left(x_1x_2+x_1x_3+x_2x_3\right)\)
\(P=\left(m+2\right)^2-6m=m^2-2m+4\)
\(P=\left(m-1\right)^2+3\ge3\)
\(\Rightarrow P_{min}=3\) khi \(m=1\)
Theo Viet ta có \(\left\{{}\begin{matrix}x_1+x_2=-\frac{3m}{2}\\x_1x_2=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(P=\left(x_1+x_2\right)^2-4x_1x_2+\left(\frac{x_1+x_2+x_1x_2\left(x_1+x_2\right)}{x_1x_2}\right)^2\)
\(P=\frac{9m^2}{4}+2\sqrt{2}+\left(\frac{-\frac{3m}{2}-\frac{\sqrt{2}}{2}\left(-\frac{3m}{2}\right)}{-\frac{\sqrt{2}}{2}}\right)^2\)
\(P=\frac{9m^2}{4}+2\sqrt{2}+\left(\frac{27-8\sqrt{2}}{4}\right)m^2\)
\(P=\left(\frac{18-9\sqrt{2}}{2}\right)m^2+2\sqrt{2}\ge2\sqrt{2}\)
\(\Rightarrow P_{min}=2\sqrt{2}\) khi \(m=0\)
PT
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(x+5\right)=m\)
\(\Leftrightarrow\left(x^2+4x+3\right)\left(x^2+4x-5\right)=m\)
\(\Leftrightarrow\left(x^2+4x-1+4\right)\left(x^2+4x-1-4\right)=m\)
\(\Leftrightarrow\left(x^2+4x-1\right)^2-16=m\)
\(\Leftrightarrow\left(x^2+4x-1\right)^2=m+16\) \(\left(DK:m\ge-16\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+4x-1=\sqrt{m+16}\left(1\right)\\x^2+4x-1=-\sqrt{m+16}\left(2\right)\end{cases}}\)
PT(1)
\(\Leftrightarrow x^2+4x-1-\sqrt{m+16}=0\)
Ta co:
\(\Delta^`=2^2-1.\left(-1-\sqrt{m+16}\right)=5+\sqrt{m+16}>0\)
\(\Rightarrow\hept{\begin{cases}x_1=-2+\sqrt{5+\sqrt{m+16}}\\x_2=-2-\sqrt{5+\sqrt{m+16}}\end{cases}}\)
PT(2)
\(\Leftrightarrow x^2+4x-1+\sqrt{m+16}=0\)
Ta lai co:
\(\Delta^`=2^2-1.\left(-1+\sqrt{m+16}\right)=5-\sqrt{m+16}\)
De PT co 4 nghiem phan biet thi PT(1) va PT(2) co 2 nghiem phan bet
Suy ra PT(2) co 2 nghiem phan biet khi
\(5-\sqrt{m+16}>0\)
\(\Leftrightarrow m< 9\)
\(\Rightarrow\hept{\begin{cases}x_3=-2+\sqrt{5-\sqrt{m+16}}\\x_4=-2-\sqrt{5-\sqrt{m+16}}\end{cases}}\)
Ta lai co:
\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_4}+\frac{1}{x_5}=\frac{x_1+x_2}{x_1x_2}+\frac{x_4+x_5}{x_4x_5}=\frac{4}{1+\sqrt{m+16}}+\frac{4}{1-\sqrt{m+16}}\text{ }=-\frac{8}{15+m}\)\(\left(DK:m\ne-15\right)\)
Ma \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=-1\)
\(\Leftrightarrow-\frac{8}{m+15}=-1\)
\(\Leftrightarrow m=-7\)
Vay de PT \(\left(x^2-1\right)\left(x+3\right)\left(x+5\right)=m\)co 4 gnhiem phan biet thoa man
\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=-1\)thi m=-7
đen ta = (2m-1)^2 - 4(m^2-1) = 4m^2 - 4m + 1 - 4m^2 + 4 = 5-4m >= 0 => m =< 5/4
p = (x1)^2 + (x2)^2 = (x1+x2)^2 - 2x1x2 = (2m-1)^2 - 2.(m^2-1) = 4m^2 - 4m + 1 - 2m^2 + 2 = 2m^2 - 4m + 2 + 1 = 2(m-1)^2 + 1 >= 1
dấu "=" xảy ra khi m = 1 (thõa mãn =< 5/4)
mậy minP = 1 khi m = 1
Phương trình 2 nghiệm phân biệt khi
\(\Delta=\left(1-m\right)^2-4\left(-m\right).1=\left(m+1\right)^2>0\)
\(\Leftrightarrow m\ne-1\)
Hệ thức Vière : \(\hept{\begin{cases}x_1+x_2=m-1\\x_1.x_2=-m\end{cases}}\)
Khi đó \(x_1\left(5-x_2\right)\ge5\left(3-x_2\right)-36\)
<=> \(-x_1x_2+5\left(x_1+x_2\right)\ge-21\)
<=> \(-\left(-m\right)+5\left(m-1\right)\ge-21\)
\(\Leftrightarrow6m\ge-16\Leftrightarrow m\ge-\frac{8}{3}\)
Kết hợp điều kiện => \(\hept{\begin{cases}m\ge-\frac{8}{3}\\m\ne-1\end{cases}}\)thì thỏa mãn bài toán
\(\Delta=\left(1-m\right)^2+4m=\left(m+1\right)^2>0\Rightarrow m\ne-1\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=-m\end{matrix}\right.\)
\(x_1\left(5-x_2\right)\ge5\left(3-x_2\right)-36\)
\(\Leftrightarrow5\left(x_1+x_2\right)-x_1x_2\ge-21\)
\(\Leftrightarrow5\left(m-1\right)+m\ge-21\)
\(\Leftrightarrow m\ge-\dfrac{8}{3}\)
Kết hợp điều kiện ban đầu ta được: \(\left\{{}\begin{matrix}m\ne-1\\m\ge-\dfrac{8}{3}\end{matrix}\right.\)