Điều kiện để phương trình trở nên có nghĩa là : \(x^2-x-6\ge0\)

Đặt : \(\sqrt{x^2-x-6}=t\left(t\ge0\right)\)

\(\Rightarrow x^2-x-18=t^2-12\left(t^2-12\ge0\right)\) 

Khi đó phương trình trở thành :

\(t^2-t-12=0\)

\(\Leftrightarrow\left(t-3\right)\left(t+4\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}t=3\left(nhận\right)\\t=-4\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow t=3\)

\(\Leftrightarrow x^2-x-6=9\)

\(\Leftrightarrow x^2-x-15=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{61}}{2}\\x_2=\dfrac{1+\sqrt{61}}{2}\end{matrix}\right.\)

\(Vậy...\)

khó thế nhỉ. lớp 5 sao giải đc

1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)

2. \(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=0\\4x+2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)

3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)