1: ĐKXĐ: x>=0
\(x+7\sqrt{x}-8=0\)
=>\(x+8\sqrt{x}-\sqrt{x}-8=0\)
=>\(\sqrt{x}\left(\sqrt{x}+8\right)-\left(\sqrt{x}+8\right)=0\)
=>\(\left(\sqrt{x}+8\right)\left(\sqrt{x}-1\right)=0\)
mà \(\sqrt{x}+8>=8\forall x>=0\)
nên \(\sqrt{x}-1=0\)
=>\(\sqrt{x}=1\)
=>x=1(nhận)
2:
a: Tọa độ A là:
\(\left\{{}\begin{matrix}x=0\\y=mx-2=m\cdot0-2=-2\end{matrix}\right.\)
vậy: A(0;-2)
b: Tọa độ B là:
\(\left\{{}\begin{matrix}y=0\\mx-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\mx=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{m}\\y=0\end{matrix}\right.\)
=>\(B\left(\dfrac{2}{m};0\right)\)
O(0;0); A(0;-2); \(B\left(\dfrac{2}{m};0\right)\)
=>\(OA=\sqrt{\left(0-0\right)^2+\left(-2-0\right)^2}=2\)
\(OB=\sqrt{\left(\dfrac{2}{m}-0\right)^2+\left(0-0\right)^2}=\sqrt{\left(\dfrac{2}{m}\right)^2}=\dfrac{2}{\left|m\right|}\)
Ta có: Ox\(\perp\)Oy
=>OA\(\perp\)OB
=>ΔOAB vuông tại O
=>\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot2\cdot\dfrac{2}{\left|m\right|}=\dfrac{2}{\left|m\right|}\)
Để \(S_{AOB}=8\) thì \(\dfrac{2}{\left|m\right|}=8\)
=>\(\left|m\right|=\dfrac{1}{4}\)
=>\(m=\pm\dfrac{1}{4}\)
