Mình muốn hóa cái acc này ạ . Cóa ai giúp mk xóa ik đc hong ạ .

Lí do : Cái acc ma ám không hỏi đc bài và còn là nguyên nhân làm máy mk bị lag

Help me

Xét :  \(x^2-8x+14\ge0\)

\(\Leftrightarrow x^2-2.x.4+16-2\ge0\)

\(\Leftrightarrow\left(x-4\right)^2-2\ge0\)

\(\Leftrightarrow\left(x-4+\sqrt{2}\right)\left(x-4-\sqrt{2}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4+\sqrt{2}\ge0\\x-4-\sqrt{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4+\sqrt{2}\le0\\x-4-\sqrt{2}\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4-\sqrt{2}\\x\ge4+\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4-\sqrt{2}\\x\le4+\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\)

Vậy \(x\ge4+\sqrt{2}\) ; \(x\le4-\sqrt{2}\) thì căn thức đc xác định.

27 , 9 , 12

\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\left(\dfrac{\sqrt{\left(x\right)^2}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\dfrac{x-1}{\sqrt{x}}\)

\(\textit{Ta có : OB = AB + OA }\)

\(\textit{hay : OB = 2 + 8 }\)

\(\textit{OB = 10 ( cm ) }\)