a: Thay m=1 vào (1), ta được:
\(x^2-5x+3\cdot1+=0\)
=>\(x^2-5x+4=0\)
=>(x-1)(x-4)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b: \(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot\left(3m+1\right)\)
\(=25-12m-4=-12m+21\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
=>-12m+21>0
=>-12m>-21
=>\(m< \dfrac{21}{12}=\dfrac{7}{4}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-5\right)}{1}=5\\x_1x_2=\dfrac{c}{a}=3m+1\end{matrix}\right.\)
\(\left|x_1^2-x_2^2\right|=15\)
=>\(\left|\left(x_1-x_2\right)\left(x_1+x_2\right)\right|=15\)
=>\(\left|5\cdot\left(x_1-x_2\right)\right|=15\)
=>\(\left|x_1-x_2\right|=3\)
=>\(\sqrt{\left(x_1-x_2\right)^2}=3\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=3\)
=>\(\sqrt{5^2-4\left(3m+1\right)}=3\)
=>\(25-12m-4=9\)
=>-12m+21=9
=>-12m=-12
=>m=1(nhận)
a. Em tự giải
b.
\(\Delta=25-4\left(3m+1\right)=21-12m>0\Rightarrow m< \dfrac{7}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=3m+1\end{matrix}\right.\)
\(\left|x_1^2-x_2^2\right|=15\)
\(\Leftrightarrow\left(x_1^2-x_2^2\right)^2=225\)
\(\Leftrightarrow\left(x_1-x_2\right)^2\left(x_1+x_2\right)^2=225\)
\(\Leftrightarrow25\left[\left(x_1+x_2\right)^2-4x_1x_2\right]=225\)
\(\Leftrightarrow25-4\left(3m+1\right)=9\)
\(\Rightarrow m=1\)
