Bài 1: Giải hệ bằng phương pháp đặt ẩn phụ
a) \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
1a) \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)( đến đây đơn giản rồi :)) )
Vậy ...
b) đặt a= 1/x và b = 1/y ( x,y khác 0)
ta có:
15a - 7b =9
4a + 9b = 35
=> a= 2, b = 3
thay vào ta có:
2 = 1/x => x = 1/2
3 = 1/y => y = 1/3
b) ĐKXĐ : x,y khác 0
Đặt 1/x = a ; 1/y = b ( a,b khác 0 )
hpt đã cho trở thành \(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)(tm)
=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=2\\\dfrac{1}{y}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)(tm)
Vậy ...
b, \(\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\)
Đặt \(\dfrac{1}{x}=t;\dfrac{1}{y}=z\)hệ phương trình tương đương
\(\left\{{}\begin{matrix}15t-7z=9\\4t+9z=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=\dfrac{9+7z}{15}\left(1\right)\\4t+9z=35\left(2\right)\end{matrix}\right.\)
Thế (1) vào (2) ta được :
\(4\left(\dfrac{9+7z}{15}\right)+9z=35\Leftrightarrow\dfrac{36+28z}{15}+\dfrac{135z}{15}=\dfrac{525}{25}\)
\(\Rightarrow163z=489\Leftrightarrow z=3\Rightarrow\dfrac{1}{y}=3\Rightarrow y=\dfrac{1}{3}\)
\(\Rightarrow t=\dfrac{9+21}{15}=2\Rightarrow\dfrac{1}{x}=2\Rightarrow x=\dfrac{1}{2}\)
Vậy \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{3}\right)\)
