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bạn đăng tách ra cho mn giúp nhé
a, Để pt có 2 nghiệm pb
\(\Delta'=1-m\ge0\Leftrightarrow m\le1\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-2\left(1\right)\\x_1x_2=m\left(2\right)\end{matrix}\right.\)
\(x_1-3x_2=0\)(3)
Từ (1) ; (3) ta có hệ \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1-3x_2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x_1=-2\\x_2=-2-x_1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{1}{2}\\x_2=-\dfrac{3}{2}\end{matrix}\right.\)
Thay vào (2) ta được \(m=\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)=\dfrac{3}{4}\)
\(b,\Delta=\left(m+5\right)^2-4\left(-m+6\right)\ge0\Leftrightarrow\left[{}\begin{matrix}m\le-7-4\sqrt{3}\\m\ge-7+4\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x1+x2=m+5\\2x1+3x2=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x1+2x2=2m+10\\2x1+3x2=13\end{matrix}\right.\)\(\)
\(\Rightarrow x2=13-2m-10=3-2m\Rightarrow x1=m+5-x2=m+5-3+2m=3m+2\)
\(x1x2=6-m\Rightarrow\left(3-2m\right)\left(3m+2\right)=6-m\Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=1\left(tm\right)\end{matrix}\right.\)
\(c,\Delta'=\left(m+1\right)^2-\left(m^2-2m+29\right)\ge0\Leftrightarrow m\ge7\)
\(\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1=2x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x2=\dfrac{2m+2}{3}\\x1=\dfrac{2\left(2m+2\right)}{3}\end{matrix}\right.\)
\(\Rightarrow x1.x2=\dfrac{\left(2m+2\right).2\left(2m+2\right)}{9}=m^2-2m+29\Leftrightarrow\left[{}\begin{matrix}m=11\left(tm\right)\\m=23\left(tm\right)\end{matrix}\right.\)
\(x^2-2\left(m+1\right)x+4m=0\)
\(\text{∆}=4\left(m+1\right)^2-16m=4\left(m-1\right)^2\)
để phương trình có 2 nghiệm phân biệt:
\(\Leftrightarrow\left(m-1\right)^2>0\Leftrightarrow m\ne1\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2\left(m+1\right)+2\left(m-1\right)}{2}=2m\\x_2=\dfrac{2\left(m+1\right)-2\left(m-1\right)}{2}=2\end{matrix}\right.\)
Ta có:
\(x_1=-3x_2\)
\(\Rightarrow2m=-6\Rightarrow m=-3\left(TM\right)\)
Vậy ...
c) Ta có: \(\text{Δ}=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot\left(2m+1\right)\)
\(=\left(-2m-2\right)^2-4\left(2m+1\right)\)
\(=4m^2+8m+4-8m-4\)
\(=4m^2\ge0\forall m\)
Do đó, phương trình luôn có nghiệm
Áp dụng hệ thức Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m+1\right)}{1}=2m+2\\x_1\cdot x_2=2m+1\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1-2x_2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x_2=2m-1\\x_1=2m+2+x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{2m-1}{3}\\x_1=2m+3+\dfrac{2m-1}{3}=\dfrac{8m+8}{3}\end{matrix}\right.\)
Ta có: \(x_1\cdot x_2=2m+1\)
\(\Leftrightarrow\dfrac{2m-1}{3}\cdot\dfrac{8m+8}{3}=2m+1\)
\(\Leftrightarrow\left(2m-1\right)\left(8m+8\right)=9\left(2m+1\right)\)
\(\Leftrightarrow16m^2+16m-8m-8-18m-9=0\)
\(\Leftrightarrow16m^2-10m-17=0\)
\(\text{Δ}=\left(-10\right)^2-4\cdot16\cdot\left(-17\right)=1188\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}m_1=\dfrac{10-6\sqrt{33}}{32}\\m_2=\dfrac{10+6\sqrt{33}}{32}\end{matrix}\right.\)
\(x^2-2\left(2m+1\right)x+4m^2+4m=0\)
Để pt có hai ng pb\(\Leftrightarrow\Delta>0\)
\(\Leftrightarrow4>0\left(lđ\right)\)
\(\Rightarrow\)Pt luôn có hai ng pb với mọi m
\(\left\{{}\begin{matrix}x_1=\dfrac{2\left(2m+1\right)+\sqrt{4}}{2}=2m+2\\x_2=\dfrac{2\left(2m+1\right)-\sqrt{4}}{2}=2m\end{matrix}\right.\)
Có \(\left|x_1-x_2\right|=x_1+x_2\)
\(\Leftrightarrow\left|2m+2-2m\right|=2m+2+2m\)
\(\Leftrightarrow2=4m+2\)
\(\Leftrightarrow m=0\)
Vậy...
\(\text{Δ}=\left(4m+1\right)^2-8\left(m-4\right)\)
\(=16m^2+8m+1-8m+32\)
\(=16m^2+33>0\)
Do đó: Phương trình luôn có hai nghiệm phân biệt
Ta có: \(\left|x_1-x_2\right|=17\)
\(\Leftrightarrow\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=17\)
\(\Leftrightarrow\sqrt{\left(4m+1\right)^2-4\cdot2\cdot\left(m-4\right)}=17\)
\(\Leftrightarrow\sqrt{16m^2+8m+1-8m+32}=17\)
\(\Leftrightarrow16m^2+33=289\)
=>m=4 hoặc m=-4