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Bài 1 : Ta xét : \(\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{5}{10}\) hay \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}\)
Nên phương trình có vô số nghiệm .
Mà \(2x+3y=5\Rightarrow x=\dfrac{5-3y}{2}\)
Vậy \(y\in R\) và \(x=\dfrac{5-3y}{2}\)
Bài 2 : \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
Đặt \(\dfrac{x}{x+1}=a\) và \(\dfrac{1}{y+4}=b\) Khi đó hệ trở thành :
\(\left\{{}\begin{matrix}3a-2b=4\\2a-5b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6a-4b=8\\6a-15b=27\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11b=-19\\6a-4b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{19}{11}\\a=\dfrac{2}{11}\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}a=\dfrac{2}{11}\\b=-\dfrac{19}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{2}{11}\\\dfrac{1}{y+4}=-\dfrac{19}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11x=2x+2\\-19y-76=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=2\\-19y=87\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{2}{9};-\dfrac{87}{19}\right)\)
Còn lại tương tự
ĐKXĐ: \(x,y\ne0\)
\(\left\{{}\begin{matrix}4x+2y=5xy\\\dfrac{2}{x}-\dfrac{5}{y}=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x+2y=5xy\\2y-5x=-4xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9x=9xy\\2y-5x=-4xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\2-5x=-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\) ( tmđk)
Vậy hệ phương trình đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(2;1\right)\)
a: \(\left\{{}\begin{matrix}3x-2y=1\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-4y=2\\2x+4y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x=5\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\2y=3x-1=\dfrac{15}{8}-1=\dfrac{7}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=\dfrac{7}{16}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}4x-3y=1\\-x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-3y=1\\-4x+8y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1+2y=-1+2=1\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{4}{3}y=1\\\dfrac{1}{2}x-\dfrac{3}{4}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=3\\2x-3y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{41}{14}\\y=-\dfrac{5}{7}\end{matrix}\right.\)
ĐK: \(x.y\ne0\)
\(\dfrac{9y}{4x}=\dfrac{4x}{y}\Leftrightarrow9y^2=16x^2\Leftrightarrow\left[{}\begin{matrix}4x=3y\\4x=-3y\end{matrix}\right.\)
TH1: \(4x=3y\) thay vào pt đầu:
\(3y+\dfrac{9y}{4}=120\Leftrightarrow\dfrac{21y}{4}=120\Rightarrow\left\{{}\begin{matrix}y=\dfrac{160}{7}\\x=\dfrac{120}{7}\end{matrix}\right.\)
TH2: \(4x=-3y\) thay vào pt đầu:
\(-3y+\dfrac{9y}{4}=120\Leftrightarrow\dfrac{-3y}{4}=120\Rightarrow\left\{{}\begin{matrix}y=-160\\x=120\end{matrix}\right.\)