2: \(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(m-3\right)\)
\(=9-4m+12=-4m+21\)
Để phương trình có hai nghiệm thì Δ>=0
=>-4m+21>=0
=>-4m>=-21
=>\(m< =\dfrac{21}{4}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=m-3\end{matrix}\right.\)
\(\left|x_1-x_2\right|=1\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=1\)
=>\(\sqrt{3^2-4\left(m-3\right)}=1\)
=>\(9-4\left(m-3\right)=1\)
=>4(m-3)=8
=>m-3=2
=>m=5(nhận)
Bài 1:
ĐKXĐ: y>2
\(\left\{{}\begin{matrix}3\left|x-1\right|+\dfrac{2}{\sqrt{y-2}}=5\\4\left|1-x\right|+\dfrac{3}{\sqrt{y-2}}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}9\left|x-1\right|+\dfrac{6}{\sqrt{y-2}}=15\\8\left|x-1\right|+\dfrac{6}{\sqrt{y-2}}=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=1\\4\left|x-1\right|+\dfrac{3}{\sqrt{y-2}}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1\in\left\{1;-1\right\}\\\dfrac{3}{\sqrt{y-2}}=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{2;0\right\}\\y-2=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{2;0\right\}\\y=3\left(nhận\right)\end{matrix}\right.\)
