Câu a : Tự vẽ .
Câu b : Tọa độ giao điểm của A và B là nghiệm phương trình :
\(2x^2=x+1\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\) và \(\left(x;y\right)=\left(-\frac{1}{2};\frac{1}{2}\right)\)
Độ dài đoạn thẳng AB là :
\(AB=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2}\)
\(=\sqrt{\left(1+\frac{1}{2}\right)^2+\left(2-\frac{1}{2}\right)^2}\)
\(=\frac{3\sqrt{2}}{2}\)
1, \(A=\frac{5}{13}+\frac{5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(-\frac{20}{41}+-\frac{21}{41}\right)+\frac{5}{7}\)
\(=1+\left(-1\right)+\frac{5}{7}=0+\frac{5}{7}=\frac{5}{7}\)
2,\(B=\frac{5}{7}.\frac{2}{11}+\frac{5}{7}.\frac{12}{11}-\frac{5}{7}.\frac{7}{11}\)
\(=\frac{5}{7}\left(\frac{2}{11}+\frac{12}{11}-\frac{7}{11}\right)=\frac{5}{7}.\frac{7}{11}=\frac{5}{11}\)
3,\(C=-\frac{2}{3}+-\frac{5}{7}+\frac{2}{3}+-\frac{2}{7}\)
\(=\left(-\frac{2}{3}+\frac{2}{3}\right)+\left(-\frac{5}{7}+-\frac{2}{7}\right)=0+-1=-1\)
4,\(H=-\frac{5}{7}.\frac{2}{11}+-\frac{5}{7}.\frac{9}{11}=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)=-\frac{5}{7}.1=-\frac{5}{7}\)
5,\(N=-\frac{5}{13}+\frac{5}{7}+\frac{20}{41}+-\frac{8}{13}+\frac{21}{41}\)
\(=\left(-\frac{5}{13}+-\frac{8}{13}\right)+\left(\frac{20}{41}+\frac{21}{41}\right)+\frac{5}{7}=-1+1+\frac{5}{7}=0+\frac{5}{7}=\frac{5}{7}\)
6, \(E=\frac{5}{7}.\frac{12}{11}+\frac{5}{7}.\frac{12}{11}-\frac{5}{7}.\frac{17}{11}=\frac{5}{7}.\left(\frac{12}{11}+\frac{12}{11}-\frac{17}{11}\right)=\frac{5}{7}.\frac{7}{11}=\frac{5}{11}\)
