a) 2x2 +x - 6 = 0
<=> (2x-3)(x+2) = 0
\(< =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
vậy ...
b) \(\left\{{}\begin{matrix}2x+3y=1\\x-y=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x+3y=1\left(1\right)\\x=y+3\left(2\right)\end{matrix}\right.\)
thay (2) vào (1) => 2(y+3)+3y=1
<=>5y = -5 <=> y = -1 => x= 2
vậy ...
a.
\(\Delta=1-4.2.\left(-6\right)=49>0\)
Phương trình đã cho có 2 nghiệm
\(\left[{}\begin{matrix}x_1=\dfrac{-1-\sqrt{49}}{2.2}=-2\\x_2=\dfrac{-1+\sqrt{49}}{2.2}=\dfrac{3}{2}\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=1\\3x-3y=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=1\\5x=10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\dfrac{1-2x}{3}\\x=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
a.\(2x^2+x-6=0\\ \Delta=b^2-4.a.c\)
=1\(^2\)-4.2.(-6)=1+48=49
\(\sqrt{\Delta}=\sqrt{49}=7\)
vậy pt có 2 nghiệm phân biệt
x\(_1\)=\(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+7}{4}\)=\(\dfrac{3}{2}\)
x\(_2=\)\(\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-7}{4}\)=-2
