e, \(\left(m-1\right)x^2-\left(m+2\right)x-2=0\)
\(\Delta=\left[-\left(m+2\right)\right]^2-4\left(-2\right)\left(m-1\right)\)
\(=m^2+4m+4+8m-8=m^2+12m-4\)
pt có 1 nghiệm \(< =>\Delta=0< =>m^2+12m-4=0\)
\(=>\Delta1=12^2-4\left(-4\right)=160>0\)
\(=>\left[{}\begin{matrix}m1=\dfrac{-12+\sqrt{160}}{2}=-6+2\sqrt{10}\\m2=\dfrac{-12-\sqrt{160}}{2}=-6-2\sqrt{10}\end{matrix}\right.\)
f, \(x^2-2x+m-5=0\)
\(\Delta=\left(-2\right)^2-4\left(m-5\right)=4-4m+20=24-4m\)
pt có nghiệm \(< =>\Delta\ge0< =>24-4m\ge0< =>m\le6\)
theo vi ét\(=>\left\{{}\begin{matrix}x1+X2=2\left(1\right)\\x1x2=m-5\left(2\right)\end{matrix}\right.\)
\(=>2x1-x2=3\)(3)
(1)(3)=> hệ \(\left\{{}\begin{matrix}x1+x2=2\\2x1-x2=3\end{matrix}\right.=>\left\{{}\begin{matrix}x1=\dfrac{5}{3}\\x2=\dfrac{1}{3}\end{matrix}\right.\)(4)
thế (4) vào(2)\(=>m-5=\dfrac{1}{3}.^{ }\dfrac{5}{3}=>m=\dfrac{5}{9}+5=\dfrac{50}{9}\left(tm\right)\)
\(\)
g, \(x^2-2\left(m-1\right)x+m^2-m+2=0\)
\(\Delta=\left[-2\left(m-1\right)\right]^2-4\left(m^2-m+2\right)\)
\(=4m^2-8m+4-4m^2+4m-8=-4m-4\)
pt có nghiệm \(< =>-4m-4\ge0< =>m\le-1\)
theo vi ét\(=>\left\{{}\begin{matrix}x1+x2=2m-2\\x1x2=m^2-m+2\end{matrix}\right.\)
có \(x1^2+x2^2=7< =>\left(x1+x2\right)^2-2x1x2-7=0\)
\(< =>\left(2m-2\right)^2-2\left(m^2-m+2\right)-7=0\)
\(< =>4m^2-8m+4-2m^2+2m-4-7=0\)
\(< =>2m^2-6m-7=0\)
\(=>\Delta1=\left(-6\right)^2-4\left(-7\right)2=92>0\)
\(=>\left[{}\begin{matrix}m1=\dfrac{6+\sqrt{92}}{2.2}=\dfrac{3+\sqrt{23}}{2}\left(loai\right)\\m2=\dfrac{6-\sqrt{92}}{2.2}=\dfrac{3-\sqrt{23}}{2}\left(loai\right)\end{matrix}\right.\)
vậy m\(\in\phi\)
g) Ta có: \(x^2-2\left(m-1\right)x+m^2-m+2=0\)
a=1; b=-2m+2; \(c=m^2-m+2\)
\(\Delta=b^2-4ac\)
\(=\left(-2m+2\right)^2-4\cdot1\cdot\left(m^2-m+2\right)\)
\(=4m^2-8m+4-4m^2+4m-8\)
\(=-4m-4\)
Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)
\(\Leftrightarrow m< -1\)
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=2m-2\\x_1\cdot x_2=\dfrac{c}{a}=m^2-m+2\end{matrix}\right.\)
Ta có: \(x_1^2+x_2^2=7\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=7\)
\(\Leftrightarrow\left(2m-2\right)^2-2\cdot\left(m^2-m+2\right)=7\)
\(\Leftrightarrow4m^2-8m+4-2m^2+2m-4=7\)
\(\Leftrightarrow2m^2-6m-7=0\)
\(\Delta=\left(-6\right)^2-4\cdot2\cdot\left(-7\right)=36+56=92\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}m_1=\dfrac{6-2\sqrt{23}}{4}=\dfrac{3-\sqrt{23}}{2}\left(loại\right)\\m_2=\dfrac{6+2\sqrt{23}}{4}=\dfrac{3+\sqrt{23}}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(m\in\varnothing\)
