Bài III
\(\left\{{}\begin{matrix}\dfrac{8}{x-3}+\dfrac{1}{2\left|y\right|-3}=5\\\dfrac{12}{x-3}+\dfrac{3}{2\left|y\right|-3}=9\end{matrix}\right.\left(Đk:x\ne3;y\ne\pm\dfrac{3}{2}\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}+\dfrac{3}{2\left|y\right|-3}=15\\\dfrac{24}{x-3}+\dfrac{6}{2\left|y\right|-3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2\left|y\right|-3}=3\\\dfrac{12}{x-3}+\dfrac{3}{2\left|y\right|-3}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|y\right|-3=1\\\dfrac{12}{x-3}+\dfrac{3}{1}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|y\right|=4\\\dfrac{12}{x-3}=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=2\\x-3=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\pm2\\x=5\end{matrix}\right.\)
Vậy: ....
Bài III:
2: Phương trình hoành độ giao điểm là:
\(x^2=3x-2\)
=>\(x^2-3x+2=0\)
=>(x-1)(x-2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Thay x=1 vào (P), ta được: \(y=1^2=1\)
Thay x=2 vào (P), ta được: \(y=2^2=4\)
Vậy: A(1;1); B(2;4)
Ta có: O(0;0); A(1;1); B(2;4)
\(OA=\sqrt{\left(1-0\right)^2+\left(1-0\right)^2}=\sqrt{1+1}=\sqrt{2}\)
\(OB=\sqrt{\left(2-0\right)^2+\left(4-0\right)^2}=2\sqrt{5}\)
\(AB=\sqrt{\left(2-1\right)^2+\left(4-1\right)^2}=\sqrt{10}\)
Xét ΔOAB có \(cosAOB=\dfrac{OA^2+OB^2-AB^2}{2\cdot OA\cdot OB}=\dfrac{20+2-10}{2\cdot2\sqrt{5}\cdot\sqrt{2}}=\dfrac{3}{\sqrt{10}}\)
=>\(sinAOB=\sqrt{1-cos^2AOB}=\dfrac{1}{\sqrt{10}}\)
Diện tích tam giác AOB là:
\(S_{AOB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB\)
\(=\dfrac{1}{2}\cdot\sqrt{2}\cdot2\sqrt{5}\cdot\dfrac{1}{\sqrt{10}}=1\)
