Question

[Trần Đức Cao Phát]

Giải hệ phương trình sau:

\(\left\{{}\begin{matrix}\dfrac{210}{x}-\dfrac{210}{y}=\dfrac{7}{4}\\4x+\dfrac{9}{4}y=210\end{matrix}\right.\)

Answer 1

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: x<>0 và y<>0

\(\left\{{}\begin{matrix}\dfrac{210}{x}-\dfrac{210}{y}=\dfrac{7}{4}\\4x+\dfrac{9}{4}y=210\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{30}{x}-\dfrac{30}{y}=\dfrac{1}{4}\\16x+9y=840\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}30\left(\dfrac{1}{x}-\dfrac{1}{y}\right)=\dfrac{1}{4}\\16x=840-9y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{120}\\x=\dfrac{840-9y}{16}\end{matrix}\right.\)

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{120}\)

=>\(\dfrac{16}{840-9y}-\dfrac{1}{y}=\dfrac{1}{120}\)

=>\(\dfrac{16y-840+9y}{y\left(840-9y\right)}=\dfrac{1}{120}\)

=>\(y\left(840-9y\right)=120\left(25y-840\right)\)

=>\(-9y^2+840y-3000y+100800=0\)

=>\(-9y^2-2160y+100800=0\)

=>\(y^2+240y-11200=0\)

=>\(\left[{}\begin{matrix}y=40\left(nhận\right)\\y=-280\left(nhận\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{840-9\cdot40}{16}=\dfrac{840-360}{16}=30\left(nhận\right)\\x=\dfrac{840-9\cdot\left(-280\right)}{16}=210\left(nhận\right)\end{matrix}\right.\)

Answer 2

[Quản Văn Mạnh]

0