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a) \(\sqrt{x-8}\) xác định khi
\(x-8\ge0\Leftrightarrow x\ge8\)
b) \(\sqrt{3x+1}\) xác định khi
\(3x+1\ge0\Leftrightarrow3x\ge-1\Leftrightarrow x\le-\dfrac{1}{3}\)
c) \(\sqrt{x^2+1}\)
Ta có: \(x^2\ge0\Rightarrow x^2+1\ge0\)
Vậy biểu thức được xác định với mọi x
d) \(\sqrt{\left(x-6\right)\left(x+3\right)}\)
Xác định khi
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-6\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-6< 0\\x+3\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\ge6\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 6\\x< -3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge6\\x< -3\end{matrix}\right.\)
e) \(\sqrt{\dfrac{-2}{x-5}}\) xác định khi
\(\left\{{}\begin{matrix}\dfrac{-2}{x-5}\ge0\\x-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ne5\end{matrix}\right.\)
\(\Leftrightarrow x< 5\)
f) \(\dfrac{4}{\sqrt{x+3}}\) xác định khi
\(x+3>0\)
\(\Leftrightarrow x>-3\)
g) \(\dfrac{6x-2}{\sqrt{x}-3}\)
Xác định khi:
\(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
h) \(\sqrt{x^2-16}=\sqrt{\left(x+4\right)\left(x-4\right)}\)
Xác định khi
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+4< 0\\x-4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+4\ge0\\x-4\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< -4\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge-4\\x\ge4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x\ge4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2,\Leftrightarrow\left\{{}\begin{matrix}20x+25y-10xy=0\\20x-30y+xy=0\end{matrix}\right.\Leftrightarrow55y-11xy=0\\ \Leftrightarrow11y\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}y=0\\x=5\end{matrix}\right.\)
Với \(y=0\Leftrightarrow4x+0=0\Leftrightarrow x=0\)
Với \(x=5\Leftrightarrow20+5y=10y\Leftrightarrow y=4\)
Vậy \(\left(x;y\right)=\left\{\left(0;0\right);\left(5;4\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\sqrt{4-x^2}=2\)
\(\Leftrightarrow4-x^2=\left(2-x\right)^2\)
\(\Leftrightarrow4-x^2=4-8x+x^2\)
\(\Leftrightarrow4-x^2-4+8x-x^2=0\)
\(\Leftrightarrow8x-2x^2=0\)
\(\Leftrightarrow2x\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(x+\sqrt{1-x^2}=1\)
\(\Leftrightarrow1-x^2=\left(1-x\right)^2\)
\(\Leftrightarrow1-x^2=1-2x+x^2\)
\(\Leftrightarrow1-x^2-1+2x-x^2=0\)
\(\Leftrightarrow2x-2x^2=0\)
\(\Leftrightarrow2x\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Bài 2:
a: Xét tứ giác ABOC có
\(\widehat{ABO}+\widehat{ACO}=180^0\)
Do đó: ABOC là tứ giác nội tiếp