Question

giải phương trình

c)\(\begin{cases} 3x+5y=1\\ 2x-y=-8 \end{cases} \)d)\(\begin{cases} \dfrac{1}{x}-\dfrac{1}{y}=1\\ \dfrac{3}{x}+\dfrac{4}{y}=5 \end{cases} \)

Answer 1

\(c,\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+10y=2\\6x-3y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13y=26\\6x-3y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\6x-3.2=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)

\(d,\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\left(I\right)\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\left(x\ne0\right)\\\dfrac{1}{y}=b\left(y\ne0\right)\end{matrix}\right.\)

\(\left(I\right)\Rightarrow\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a-3b=3\\3a+4b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-7b=-2\\3a+4b=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\3a+4.\dfrac{2}{7}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\a=\dfrac{9}{7}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{2}{7}\Leftrightarrow x=\dfrac{7}{2}\\\dfrac{1}{y}=\dfrac{9}{7}\Leftrightarrow y=\dfrac{7}{9}\end{matrix}\right.\)

Answer 2

c. \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+10y=2\\6x-3y=-24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=26\\2x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\left(x\ne0\right)\\\dfrac{1}{y}=b\left(y\ne0\right)\end{matrix}\right.\)

hpt \(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-3b=3\\3a+4b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7b=-2\\a-b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{2}{7}\\a=\dfrac{9}{7}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{9}{7}\\\dfrac{1}{y}=\dfrac{2}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{9}\\y=\dfrac{7}{2}\end{matrix}\right.\)