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Bài 10:
a: \(\overrightarrow{AB}+\overrightarrow{BO}+\overrightarrow{OA}\)
\(=\overrightarrow{AO}+\overrightarrow{OA}=\overrightarrow{0}\)
b: \(\overrightarrow{OA}+\overrightarrow{BC}+\overrightarrow{DO}+\overrightarrow{CD}\)
\(=\overrightarrow{OA}+\overrightarrow{DO}+\overrightarrow{BD}\)
\(=\overrightarrow{OA}+\overrightarrow{BO}=\overrightarrow{BA}\)
ĐKXĐ: \(x>3\)
\(\Leftrightarrow2x+2\sqrt{x-3}\sqrt{x+3}=\dfrac{4\left(x+3\right)}{\left(x-3\right)^2}\)
\(\Leftrightarrow\left(\sqrt{x+3}+\sqrt{x-3}\right)^2=\dfrac{4\left(x+3\right)}{\left(x-3\right)^2}\)
\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-3}=\dfrac{2\sqrt{x+3}}{x-3}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x+3}-\sqrt{x-3}}=\dfrac{\sqrt{x+3}}{x-3}\)
\(\Leftrightarrow3x-9=x+3-\sqrt{x^2-9}\)
\(\Leftrightarrow\sqrt{x^2-9}=12-2x\) (\(x\le6\))
\(\Leftrightarrow x^2-9=144-48x+4x^2\)
\(\Leftrightarrow3x^2-48x+153=0\)
\(\Leftrightarrow x=8-\sqrt{13}\)
1b)
Song song => (d): x-y +a =0
Vì d đi qua C(2;-2) => 2- (-2)+a=0
<=>a=4
=> d: x-y+4=0
ĐKXĐ: \(x;y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=6\\xy\left(x+y\right)=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y+2\sqrt{xy}\right)=36\\xy\left(x+y\right)=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)+2xy\sqrt{xy}=36\\xy\left(x+y\right)=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\sqrt{xy}=8\\xy\left(x+y\right)=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=4\\xy\left(x+y\right)=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=4\\x+y=5\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(1;4\right);\left(4;1\right)\)