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\(\Delta=a^2-4\left(b+2\right)>0\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-a\\x_1x_2=b+2\end{matrix}\right.\) (1)
\(\left\{{}\begin{matrix}x_1-x_2=4\\\left(x_1-x_2\right)^3+3x_1x_2\left(x_1-x_2\right)=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=4\\64+12x_1x_2=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=4\\x_1x_2=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=3\\x_2=-1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x_1=1\\x_2=-3\end{matrix}\right.\)
Thế vào (1) để tìm a; b
Để pt có hai nghiệm pb \(\Leftrightarrow\Delta>0\)\(\Leftrightarrow4-4\left(m-1\right)>0\)\(\Leftrightarrow2>m\)
Theo viet có:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-1\end{matrix}\right.\)
Có \(x_1^2+x_2^2-3x_1x_2=2m^2+\left|m-3\right|\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-5x_1x_2=2m^2+\left|m-3\right|\)
\(\Leftrightarrow4-5\left(m-1\right)=2m^2+\left|m-3\right|\)
\(\Leftrightarrow2m^2+\left|m-3\right|-9+5m=0\) (1)
TH1: \(m\ge3\)
PT (1) \(\Leftrightarrow2m^2+m-3-9+5m=0\)
\(\Leftrightarrow2m^2+6m-12=0\)
Do \(m\ge3\Rightarrow\left\{{}\begin{matrix}6m-12\ge6>0\\2m^2>0\end{matrix}\right.\)
\(\Rightarrow2m^2+6m-12>0\)
=>Pt vô nghiệm
TH2: \(m< 3\)
PT (1)\(\Leftrightarrow2m^2-\left(m-3\right)-9+5m=0\)
\(\Leftrightarrow2m^2+4m-6=0\) \(\Leftrightarrow2m^2-2m+6m-6=0\)
\(\Leftrightarrow2m\left(m-1\right)+6\left(m-1\right)=0\)\(\Leftrightarrow\left(2m+6\right)\left(m-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-3\\m=1\end{matrix}\right.\) (Thỏa)
Vậy...
a: Thay m=-5 vào (1), ta được:
\(x^2+2\left(-5+1\right)x-5-4=0\)
\(\Leftrightarrow x^2-8x-9=0\)
=>(x-9)(x+1)=0
=>x=9 hoặc x=-1
b: \(\text{Δ}=\left(2m+2\right)^2-4\left(m-4\right)=4m^2+8m+4-4m+16=4m^2+4m+20>0\)
Do đó: Phương trình luôn có hai nghiệm phân biệt
\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=-3\)
\(\Leftrightarrow x_1^2+x_2^2=-3x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)
\(\Leftrightarrow\left(2m+2\right)^2+m-4=0\)
\(\Leftrightarrow4m^2+9m=0\)
=>m(4m+9)=0
=>m=0 hoặc m=-9/4
\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)
\(\Rightarrow\Delta>0\Leftrightarrow\left(m+1\right)^2-4\left(m+4\right)>0\Leftrightarrow\left[{}\begin{matrix}m< -3\\m>5\end{matrix}\right.\)\(\left(2\right)\)
\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)
\(x1\ne0\) \(\Rightarrow0< x1< x2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)
\(\left(2\right)\left(3\right)\Rightarrow m>5\)
\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)
\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)
\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)
\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow m\in\phi\)
Để pt có 2 nghiệm pb
\(\left(m+1\right)^2-4\left(m+4\right)=m^2+2m+1-4m-16\)
\(=m^2-2m-15>0\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m+4\end{matrix}\right.\)
Ta có : \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=12\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=12\)
Thay vào ta được \(m+1+2\sqrt{m+4}=12\Leftrightarrow2\sqrt{m+4}=11-m\)đk : m >= -4
\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)
\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)
a, \(\Delta=m^2-4\left(-4\right)=m^2+16\)> 0
Vậy pt luôn có 2 nghiệm pb
b, Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=-4\end{matrix}\right.\)
Ta có \(\left(x_1+x_2\right)^2-2x_1x_2=5\)
Thay vào ta được \(m^2-2\left(-4\right)=5\Leftrightarrow m^2+3=0\left(voli\right)\)
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b) phương trình có 2 nghiệm \(\Leftrightarrow\Delta'\ge0\)
\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)
\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)
\(\Leftrightarrow-4m+4\ge0\)
\(\Leftrightarrow m\le1\)
Ta có: \(x_1^2+x_1x_2+x_2^2=1\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)
Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)
\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)
\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)
\(\Leftrightarrow4m^2-10m-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)
\(x^2+\left(4m+1\right)x+2\left(m-4\right)=0\)
\(\Delta=\left(4m+1\right)^2-4\cdot1\cdot2\left(m-4\right)=16m^2+8m+1-8m+32=16m^2+33\ge33>0\forall m\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-\left(4m+1\right)+\sqrt{16m^2+33}}{2}\\x_2=\dfrac{-\left(4m+1\right)-\sqrt{16m^2+33}}{2}\end{matrix}\right.\)
Mà: \(x_2-x_1=17\)
\(\Leftrightarrow\dfrac{-\left(4m+1\right)-\sqrt{16m^2+33}}{2}-\dfrac{-\left(4m+1\right)+\sqrt{16m^2+33}}{2}=17\)
\(\Leftrightarrow\dfrac{-\left(4m+1\right)-\sqrt{16m^2+33}+\left(4m+1\right)-\sqrt{16m^2+33}}{2}=17\)
\(\Leftrightarrow\dfrac{-2\sqrt{16m^2+33}}{2}=17\)
\(\Leftrightarrow\sqrt{16m^2+33}=-17< 0\)
Vậy không có m thỏa mãn
\(x^2+ax+b+1=0\)
\(\Delta=a^2-4b-4\)
Để pt có 2 nghiệm pb \(\Leftrightarrow\Delta>0\Leftrightarrow a^2-4b-4>0\)
Theo hệ thức Vi-et ta có: \(\hept{\begin{cases}x_1+x_2=-a\\x_1.x_2=b+1\end{cases}}\)
Ta có: \(\hept{\begin{cases}x_1-x_2=3\\x_1^3-x_2^3=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1-x_2=3\\\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1-x_2=3\\x_1^2+x_1x_2+x_2^2=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1-x_2=3\\\left(x_1-x_2\right)^2+3x_1x_2=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1-x_2=3\\x_1x_2=-2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1=3+x_2\\\left(3+x_2\right)x_2=-2\left(1\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow x_2^2+3x_2+2=0\)
\(\Delta=1\)
\(\Rightarrow\)pt có 2 nghiệm pb \(\orbr{\begin{cases}x_2=\frac{-3+1}{2}=-1\Rightarrow x_1=2\\x_2=\frac{-3-1}{2}=-2\Rightarrow x_1=1\end{cases}}\)
TH1: \(x_1=2;x_2=-1\)
\(\Rightarrow\hept{\begin{cases}a=-1\\b=-3\end{cases}}\)( LOẠI vì a^2 -4b-4 <0 )
TH2: \(x_1=1;x_2=-2\)
\(\Rightarrow\hept{\begin{cases}a=1\\b=-3\end{cases}}\)( tm )
VẬY ...