a)\(\left\{{}\begin{matrix}\left(x+3\right)\left(y-5\right)=xy\\\left(x-2\right)\left(y+5\right)=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-5x+3y-15=xy\\xy+5x-2y-10=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+3y-15=0\\5x+2y-10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=15\left(1\right)\\5x+2y=10\left(2\right)\end{matrix}\right.\)\(\left(1\right)-\left(2\right)=-y=-25\Leftrightarrow y=25\)
thay y = 25 vào \(\left(2\right)\), ta có: \(5x-2.25=10\Leftrightarrow x=12\)
Vậy hệ phương trình có nghiệm (x; y) là (12; 25)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{1}{6x}+\dfrac{1}{5y}=\dfrac{2}{15}\end{matrix}\right.\)
pt 2 nhân cho 30 ta đc hệ mới là
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{5}{x}+\dfrac{6}{y}=4\end{matrix}\right.\)
đến đây ta chỉ việc đặt 1/x=a; 1/y =b
phần còn lại tương đối dễ dàng để bạn thử sức
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{1}{6x}+\dfrac{1}{5y}=\dfrac{2}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{4}\\\dfrac{30}{6x}+\dfrac{30}{5y}=\dfrac{2.30}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{4}-\dfrac{1}{y}\\\dfrac{5}{x}+\dfrac{6}{y}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{4}-\dfrac{1}{y}\\5\left(\dfrac{3}{4}-\dfrac{1}{y}\right)+\dfrac{6}{y}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{4}-\dfrac{1}{y}\\\dfrac{15}{4}-\dfrac{5}{y}+\dfrac{6}{y}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{4}-\dfrac{1}{y}\\\dfrac{1}{y}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy hệ phương trình có ngiệm (x; y) là (2; 4)
