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Để pt có 2 nghiệm
\(\Leftrightarrow\left\{{}\begin{matrix}m-1\ne0\\\Delta'=\left(m+1\right)^2-m\left(m-1\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne1\\3m+1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne1\\m\ge-\frac{1}{3}\end{matrix}\right.\)
Khi đó theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=\frac{2\left(m+1\right)}{m-1}\\x_1x_2=\frac{m}{m-1}\end{matrix}\right.\)
\(\left|x_1-x_2\right|\ge2\Leftrightarrow\left(x_1-x_2\right)^2\ge4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2\ge4\)
\(\Leftrightarrow4\left(\frac{m+1}{m-1}\right)^2-\frac{4m}{m-1}\ge4\)
\(\Leftrightarrow\left(1+\frac{2}{m-1}\right)^2-\left(1+\frac{1}{m-1}\right)-1\ge0\)
Đặt \(\frac{1}{m-1}=t\)
\(\Rightarrow\left(2t+1\right)^2-\left(t+1\right)-1\ge0\)
\(\Leftrightarrow4t^2+3t-1\ge0\Rightarrow\left[{}\begin{matrix}t\ge\frac{1}{4}\\t\le-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{m-1}\ge\frac{1}{4}\\\frac{1}{m-1}\le-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{5-m}{m-1}\ge0\\\frac{m}{m-1}\le0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}1< m\le5\\0\le m< 1\end{matrix}\right.\)
\(\Rightarrow m_{max}=5\)
a/ Ta có : △' = (-2)2-(m+3)
=4-m-3 = 1-m
De ptr co 2 nghiem x1 va x2 thì △' ≥0
=>1-m≥0 =>m≤1
Theo Viei{ x1+x2=4 ; x1x2=m+3
Ta co: |x1-x2|=2 ⇔(x1-x2)2=4
⇔(x1+x2)2-4x1x2=4
⇔42-4(m+3)=4
⇔m=0 (TM)
b/ Ta co: △ = (m-1)2-4(m+6)
=m2-6m-23 De ptr co 2 nghiem x1 , x2 thi △≥ 0
=> m2-6m-23≥0 (*)
Theo viet { x1+x2=1-m ; x1x2=m+6
db <=> ( x1+x2)2-2x1x2=10
⇔ (1-m)2-2(m+6)=10
⇔ m2-4m -21 =0
⇔[m=7 ; m=-3
Thay vao (*) =>m=7 (loai) ; m=-3 (tm)
c/ Ta co :△' = (-m)2-(3m-2)
= m2-3m+2
De ptr co 2 nghiem x1 , x2 thi : △' ≥0
⇔m2-3m+2≥0 (*)
Theo viet { x1+x2=2m ; x1x2=3m-2
db <=> ( x1+x2)2-3x1x2=4
⇔ (2m)2-3(3m-2)=4
⇔ 4m2--9m+2 =0
⇔[m=2 ; m=\(\dfrac{1}{4}\)
Thay vao (*) =>m=2 (tm) ; m=\(\dfrac{1}{4}\) (tm)
d/ Ta co : △=(-3)2-4(m-2)
=17-4m
De ptr co 2 nghiem x1 , x2 thi : △ ≥0
⇔17-4m≥0
⇔m≤\(\dfrac{17}{4}\)
theo viet{ x1+x2=3 ; x1x2= m-2
⇔(x1+x2)3-3x1x2(x1+x2) =9
⇔33-3.3(m-2)=9
⇔m=4(tm)
Thế \(\hept{\begin{cases}x_1^2=2mx_1+3m\\x_2^2=2mx_2+3m\end{cases}}\) vô cái dưới là xong nha
\(x^2-4x+m+3=0\)
\(\Delta=\left(-4\right)^2-4\left(m+3\right)=4-4m\)
Pt có 2 nghiệm \(\Rightarrow\Delta>0\Leftrightarrow4-4m>0\Leftrightarrow m< 1\)
Áp dụng hệ thức Vi-ét ta có:
\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1.x_2=m+3\end{matrix}\right.\)
Theo đề bài:
\(\left|x_2-x_1\right|=2\Leftrightarrow\left(x_2-x_1\right)^2=4\)
\(\Leftrightarrow x_2^2+x_1^2-2x_1x_2=4\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4\)
\(\Leftrightarrow4^2-4\left(m+3\right)=4\)
\(\Leftrightarrow4m=0\Leftrightarrow m=0\) (t/m)
KL: m=0 thỏa mãn đề bài
\(\Delta'=m^2-4\ge0\Rightarrow m\le-2\) (do m âm)
Khi đó theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2m>0\\x_1x_2=4>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1>0\\x_2>0\end{matrix}\right.\)
\(\left(\frac{x_1}{x_2}\right)^2+\left(\frac{x_2}{x_1}\right)^2=3\Leftrightarrow\left(\frac{x_1}{x_2}\right)^2+2\left(\frac{x_1}{x_2}\right)\left(\frac{x_2}{x_1}\right)+\left(\frac{x_2}{x_1}\right)^2-2=3\)
\(\Leftrightarrow\left(\frac{x_1}{x_2}+\frac{x_2}{x_1}\right)^2=5\Leftrightarrow\frac{x_1}{x_2}+\frac{x_2}{x_1}=\sqrt{5}\) (do \(x_1;x_2>0\))
\(\Leftrightarrow x_1^2+x_2^2=\sqrt{5}x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=\sqrt{5}x_1x_2\)
\(\Leftrightarrow4m^2-8=4\sqrt{5}\)
\(\Leftrightarrow m^2=2+\sqrt{5}\)
\(\Leftrightarrow m=-\sqrt{2+\sqrt{5}}\)