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1) \(x^2-2mx+m-2=0\) (1)
pt (1) có \(\Delta'=\left(-m\right)^2-\left(m-2\right)=m^2-m+2=\left(m-\frac{1}{2}\right)^2+\frac{7}{4}>0\left(\forall m\right)\)
=> pt luôn có 2 nghiệm phân biệt x1, x2
Vi-et: \(\hept{\begin{cases}x_1+x_2=2m\\x_1x_2=m-2\end{cases}}\)\(\Rightarrow\)\(M=\frac{2x_1x_2-\left(x_1+x_2\right)}{x_1^2+x_2^2-6x_1x_2}=\frac{2x_1x_2-\left(x_1+x_2\right)}{\left(x_1+x_2\right)^2-8x_1x_2}=\frac{2m-4-2m}{\left(2m\right)^2-8m-16}\)
\(=\frac{-4}{4m^2-8m-16}=\frac{-4}{4\left(m-1\right)^2-20}\ge\frac{-4}{-20}=\frac{1}{5}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(m=1\)
xin 1slot sáng giải
Đề đúng là \(m^3-3m\) chứ bạn?
\(\Delta'=m^2-m^3-3m\ge0\)
\(\Leftrightarrow m\left(-m^2+m-3\right)\ge0\)
\(\Rightarrow m\le0\) (do \(-m^2+m-3=-\left(m-\frac{1}{2}\right)^2-\frac{11}{4}< 0;\forall m\))
b/ \(x_1^2+x_2^2\ge8\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2\ge8\)
\(\Leftrightarrow4m^2-2m^3+6m\ge8\)
\(\Leftrightarrow m^3-2m^2-3m+4\le0\)
\(\Leftrightarrow\left(m-1\right)\left(m^2-m-4\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}m\le\frac{1-\sqrt{17}}{2}\\1\le m\le\frac{1+\sqrt{17}}{2}\end{matrix}\right.\) \(\Rightarrow m\le\frac{1-\sqrt{17}}{2}\)
Hình như đề thiếu, pt: \(x^2-\left(m+1\right)x+m-2=0\)
Phương trình đã cho có nghiệm khi \(\Delta=\left(m+1\right)^2-4\left(m-2\right)=m^2-2m+9>0\)
\(\Rightarrow\) Phương trình đã cho luôn có hai nghiệm phân biệt với mọi giá trị m
Định lí Viet: \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m-2\end{matrix}\right.\)
a, Theo giả thiết ta có: \(x_1^2+x_2^2=100\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=100\)
\(\Leftrightarrow\left(m+1\right)^2-2\left(m-2\right)=100\)
\(\Leftrightarrow m^2+2m+1-2m+4=100\)
\(\Leftrightarrow m^2=95\)
\(\Leftrightarrow m=\sqrt{95}\)
b, \(P=\left|x_1-x_2\right|\)
\(P^2=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(=\left(m+1\right)^2-4\left(m-2\right)\)
\(=m^2-2m+9=\left(m-1\right)^2+8\ge8\)
\(\Rightarrow P=\left|x_1-x_2\right|\ge2\sqrt{2}\)
\(minP=2\sqrt{2}\Leftrightarrow m=1\)
\(\Delta'=\left(m+1\right)^2-\left(5m+1\right)=m^2-3m\ge0\Rightarrow\left[{}\begin{matrix}m\ge3\\m\le0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=5m+1\end{matrix}\right.\)
a.
\(S=\left(x_1+x_2\right)^2-3x_1x_2=4\left(m+1\right)^2-3\left(5m+1\right)\)
\(=4m^2-7m+1=\dfrac{7}{3}\left(m^2-3m\right)+\dfrac{5}{3}m^2+1\ge1\)
\(S_{min}=1\) khi \(\dfrac{7}{3}\left(m^2-3m\right)+\dfrac{5}{3}m^2=0\Rightarrow m=0\)
b.
\(1< x_1< x_2\Rightarrow\left\{{}\begin{matrix}\left(x_1-1\right)\left(x_2-1\right)>0\\\dfrac{x_1+x_2}{2}>1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1x_2-\left(x_1+x_2\right)+1>0\\x_1+x_2>2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5m+1-2\left(m+1\right)+1>0\\2\left(m+1\right)>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m>0\\m>-1\end{matrix}\right.\) \(\Rightarrow m>0\)
Kết hợp điều kiện delta \(\Rightarrow m\ge3\)
\(a,\Leftrightarrow\Delta\ge0\Leftrightarrow\left(2m+2\right)^2-4\left(5m+1\right)\ge0\Leftrightarrow4m^2-12m\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\le0\\m\ge3\end{matrix}\right.\)
\(vi-ét\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1x2=5m+1\end{matrix}\right.\)
\(\Rightarrow S=x1^2+x2^2-x1x2=\left(x1+x2\right)^2-3x1x2\)
\(=\left(2m+2\right)^2-3\left(5m+1\right)=4m^2-7m+1\)
\(=\left(2m\right)^2-2.2.\dfrac{7}{4}.m+\left(\dfrac{7}{4}\right)^2-\dfrac{33}{16}=\left(2m-\dfrac{7}{4}\right)^2-\dfrac{33}{16}\left(1\right)\)
\(TH1:m\ge3\Rightarrow\left(1\right)\ge\left(2.3-\dfrac{7}{4}\right)^2-\dfrac{33}{16}=16\)
\(TH2:m\le0\Rightarrow\left(1\right)\ge\left(0-\dfrac{7}{4}\right)^2-\dfrac{33}{16}=1\)
\(\Rightarrow MinS=1\Leftrightarrow m=0\left(tm\right)\)
\(b,1< x1< x2\Leftrightarrow0< x1-1< x2-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left(x1-1\right)\left(x2-1\right)>0\\x1+x2-2>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x1>1\\x2>1\end{matrix}\right.\\\left\{{}\begin{matrix}x1 < 1\\x2< 1\end{matrix}\right.\end{matrix}\right.\\2m+2-2>0\\\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}x1x2>1\\x1x2< 1\end{matrix}\right.\\m>0\\\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m>3\\m< 0\end{matrix}\right.\\\left[{}\begin{matrix}m>0\\m< 0\end{matrix}\right.\\m>0\\\end{matrix}\right.\Rightarrow m>3\)