Bài 3:
1) \(x^2-5x+6=0\) (1)
a) \(a=1;b=-5;c=6\)
b) \(\Delta=\left(-5\right)^2-4\cdot1\cdot6=1>0\)
Phương trình (1) có 2 nghiệm phân biệt
\(x_1=\dfrac{5+\sqrt{1}}{2}=3\)
\(x_2=\dfrac{5-\sqrt{1}}{2}=2\)
Vậy: ....
2) \(x^2-mx+m-1=0\)(2)
a) \(\Delta=\left(-m\right)^2-4\cdot1\cdot\left(m-1\right)=m^2-4m+4\)
\(=m^2-2\cdot m\cdot2+2^2=\left(m-2\right)^2\ge0\forall m\)
Phương trình luôn có 2 nghiệm với mọi m
b) Thay \(x=3\) (2) ta có:
\(3^2-3m+m-1=0\)
\(\Leftrightarrow-2m+8=0\)
\(\Leftrightarrow-2m=-8\)
\(\Leftrightarrow m=4\)
Theo vi-ét: \(x_1+x_2=\dfrac{-\left(-m\right)}{1}=m=4\)
\(\Rightarrow x_2=4-x_1=4-3=1\)
Bài 1:
\(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\11x-15=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3\cdot3-5\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
Vậy: ...
