Lời giải:
Để pt có hai nghiệm phân biệt thì \(\Delta'=1+2m>0\Leftrightarrow m> \frac{-1}{2}\)
a)
Áp dụng hệ thức Viete, với $x_1,x_2$ là hai nghiệm của pt:
\(\left\{\begin{matrix} x_1+x_2=2\\ x_1x_2=-2m\end{matrix}\right.\)
Khi đó: \((x_1^2+1)(x_2^2+1)=5\)
\(\Leftrightarrow (x_1x_2)^2+x_1^2+x_2^2=4\)
\(\Leftrightarrow (x_1x_2)^2+(x_1+x_2)^2-2x_1x_2=4\)
\(\Leftrightarrow 4m^2+4+4m=4\)
\(\Leftrightarrow m(m+1)=0\Rightarrow m=0\) do \(m> \frac{-1}{2}\)
b)
Ta có:
\(u=\frac{1}{x_1+1}+\frac{1}{x_2+1}=\frac{x_1+x_2+2}{(x_1+1)(x_2+1)}\)
\(=\frac{x_1+x_2+2}{x_1x_2+(x_1+x_2)+1}=\frac{2+2}{-2m+2+1}=\frac{4}{3-2m}\)
\(v=\frac{1}{x_1+1}.\frac{1}{x_2+1}=\frac{1}{(x_1+1)(x_2+1)}=\frac{1}{x_1+x_2+x_1x_2+1}=\frac{1}{2-2m+1}=\frac{1}{3-2m}\)
Do đó pt nhận \(\frac{1}{x_1+1}; \frac{1}{x_2+1}\) làm nghiệm theo định lý Viete đảo là:
\(X^2-\frac{4}{3-2m}X+\frac{1}{3-2m}=0\)
\(\Leftrightarrow (3-2m)X^2-4X+1=0\)
f(x) =x^2 -2x -2m
a) f(x) có hai nghiệm pb <=> 1 +2m > 0 => m>-1/2
P=\(\left(x_1^2+1\right)\left(x_2^2+1\right)=\left(x_1.x_2\right)^2+\left(x_1+x_2\right)^2-2x_1x_2+1\)
\(P=\left(x_1x_2-1\right)^2+\left(x_1+x_2\right)^2=\left(2m+1\right)^2+4\)
\(P=5\Leftrightarrow\left(2m+1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}2m+1=-1;m=-1\left(l\right)\\2m+1=1;m=0\left(n\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m\ge\dfrac{1}{2}\\1+2-2m\ne0\end{matrix}\right.\) <=> \(m\in[\dfrac{-1}{2};\dfrac{3}{2})U\left(\dfrac{3}{2};\infty\right)\)
\(\left\{{}\begin{matrix}\dfrac{1}{x_1+1}+\dfrac{1}{x_2+1}=\dfrac{x_1+x_2+2}{x_1x_2+\left(x_1+x_2\right)+1}=\dfrac{4}{3-2m}\\\dfrac{1}{x_1+1}.\dfrac{1}{x_2+1}=\dfrac{1}{3-2m}\end{matrix}\right.\)
phương trình cần tìm
\(g\left(x\right)=x^2-\dfrac{4}{3-2m}+\dfrac{1}{3-2m}\) \(\Leftrightarrow\left\{{}\begin{matrix}m\in[\dfrac{-1}{2};\dfrac{3}{2})U\left(\dfrac{3}{2};\infty\right)\\\left(2m-3\right)x^2+4x-1=0\end{matrix}\right.\)
