\(c.x^2+\left(x+3\right)\left(x-5\right)-9=0\\ x^2-9+\left(x+3\right)\left(x-5\right)=0\\ \left(x-3\right)\left(x+3\right)+\left(x+3\right)\left(x-5\right)=0\\ \left(x+3\right)\left(x-3+x-5\right)=0\\ \left(x-3\right)\left(2x-8\right)=0\)
\(\left[{}\begin{matrix}x-3=0\\2x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
\(d.\dfrac{\left(x+2\right)^2-3\left(x-2\right)-3x-10}{\left(x-2\right)\left(x+2\right)}=0\\ \dfrac{x^2+4x+4-3x+6-3x-10}{\left(x-2\right)\left(x+2\right)}=0\\ \dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=0\\ \dfrac{x}{x+2}=0\\ x=0\)
\(c,x^2+\left(x+3\right)\left(x-5\right)=9\)
\(\Leftrightarrow x^2+x^2-5x+3x-15-9=0\)
\(2x^2-2x-24=0\)
\(\Delta=b^2-4ac=\left(-2\right)^2-4.2.\left(-24\right)=196>0\)
\(\Rightarrow\) Pt có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{2+14}{4}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{2-14}{4}=-3\end{matrix}\right.\)
Vậy \(S=\left\{4;-3\right\}\)
\(d,\) Với \(đk:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)-3\left(x-2\right)-3x-10}{x^2-4}=0\)
\(\Leftrightarrow\dfrac{x^2+2x+2x+4-3x+6-3x-10}{x^2-4}=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\x=-2\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
vẽ hình giúp mik nha mik cảm ơn rất rất nhiều
giúp mik vs mik đag cần gấp