Bài 2:
\(Thay:x_A=1;y_A=7.vào.ĐTHS:7=a+b\left(1\right)\\ Thay:x_A=-1;y_A=-3.vào.ĐTHS:-3=-a+b\left(2\right)\\ Từ\left(1\right),\left(2\right),lập.hpt:\left\{{}\begin{matrix}a+b=7\\-a+b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right. \)
Bài 3:
a, Khi m=1 ta được pt:
\(\left\{{}\begin{matrix}x+2y=1\\2x+5y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=2\\2x+5y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-y=1\\x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x+2.\left(-1\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(3;-1\right)\)
b, Khi: x+2y=10 => x= 10-2y
\(\Leftrightarrow\left\{{}\begin{matrix}10-2y+2y=m\\2\left(10-2y\right)+5y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=10\\20-4y+5y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m=10\\y=-19\Rightarrow x=10-2y=10-2.\left(-19\right)=48\end{matrix}\right.\)
Vậy khi m=10 thì hpt có nghiệm (x;y)= (48;-19) thoả mãn x+2y=10
Bài 1:
\(Đặt:a=\dfrac{1}{x};b=\dfrac{1}{y}\left(x,y\ne0\right)\\ Hpt:\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=4\\\dfrac{4}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+3b=4\\4a-b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4a+6b=8\\4a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a-b=1\\7b=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{b+1}{4}\\b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1+1}{4}=\dfrac{1}{2}\\b=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{x}=\dfrac{1}{2}\\b=\dfrac{1}{y}=1\end{matrix}\right.\\ Vậy:x=2;y=1\Rightarrow\left(x;y\right)=\left(2;1\right)\)
