b.\(\left\{{}\begin{matrix}11x-3y=-7\\4x+15y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}55x-15y=-35\\4x+15y=-24\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}59x=-59\\11x-3y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\11.\left(-1\right)-3y=-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\-11-3y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\-3y=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{-4}{3}\end{matrix}\right.\)
vậy pt có nghiệm duy nhất(-1;\(\dfrac{-4}{3}\))
a)ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{3}{x-2}+\dfrac{x}{x+2}=\dfrac{8}{4-x^2}\)
\(\Leftrightarrow\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-8}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(3x+6+x^2-2x+8=0\)
\(\Leftrightarrow x^2+x+14=0\)(vô lý)
Vậy: \(S=\varnothing\)
