`a) x^2 - x - 6 = 0`
Ptr có: `\Delta = b^2 - 4ac = (-1)^2 - 4 . 1 . (-6) = 25 > 0`
`=>` Ptr có `2` nghiệm pb
`x_1 = [ -b + \sqrt{\Delta} ] / [ 2a ] = [ -(-1) + \sqrt{25} ] / [ 2 . 1 ] = 3`
`x_2 = [ -b - \sqrt{\Delta} ] / [ 2a ] = [ -(-1) - \sqrt{25} ] / [ 2 . 1 ] = -2`
Vậy `S = { 3 ; -2 }`
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`b) {(4x - 3y = 4),(-6x + 7y = 4):}`
`<=>{(28x - 21y = 28),(-18x + 21y = 12):}`
`<=>{(10x = 40),(4x - 3y = 4):}`
`<=>{(x = 4),(4 . 4 - 3y = 4):}`
`<=>{(x = 4),(y = 4):}`
Vậy hệ ptr có nghiệm `(x ; y ) = ( 4 ; 4 )`
\(a,x^2-x-6=0\)
\(\Delta=b^2-4ac=\left(-1\right)^2-4\left(-6\right)=25>0\)
\(\Rightarrow\)Pt có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+5}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-5}{2}=-2\end{matrix}\right.\)
Vậy \(S=\left\{3;-2\right\}\)
\(b,\left\{{}\begin{matrix}4x-3y=4\\-6x+7y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x-9y=12\\12x-14y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x-9y-12x+14y=12-\left(-8\right)\\4x-3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=20\\4x-3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\4x-3.4=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\4x-12=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\4x=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=4\end{matrix}\right.\)
Vậy hệ pt có tập nghiệm\(\left(x;y\right)=\left(4;4\right)\)
\(\left\{{}\begin{matrix}4x-3y=4\\-6x+7y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=4+6x\\4x-3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{7}+\dfrac{6x}{7}\\4x-3\left(\dfrac{4}{7}+\dfrac{6x}{7}\right)=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4+6x}{7}\\\dfrac{10x}{7}=\dfrac{40}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{6x+4}{7}\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{6.4+4}{7}\\x=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=4\end{matrix}\right.\)
\(x^2-x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
