hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 ,...
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hệ phương trình
1 ,\(\left\{{}\begin{matrix}\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{3}{2}\\3x-2y=5\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{x^2-y-6}{x}=x-2\\x+3y=8\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{x}{y}=\frac{2}{3}\\x+y=10\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\frac{y^2+2x-8}{y}=y-3\\x+y=10\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}\frac{x+1}{y-1}=5\\3\left(2x-2\right)-4\left(3x+4\right)=5\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}2x+y=4\\\left|x-2y\right|=3\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{2x}{x+1}+\frac{y}{y+1}=3\\\frac{x}{x+1}-\frac{3y}{y+1}=-1\end{matrix}\right.\)
9 , \(\left\{{}\begin{matrix}y-\left|x\right|=1\\2x-y=1\end{matrix}\right.\)
10 , \(\left\{{}\begin{matrix}\sqrt{x+3y}=\sqrt{3x-1}\\5x-y=9\end{matrix}\right.\)
\(\frac{3\left(a^2+1\right)}{a}=\frac{4\left(b^2+1\right)}{b}=\frac{5\left(c^2+1\right)}{c}\)
\(\Leftrightarrow\frac{3\left(a+b\right)\left(a+c\right)}{a}=\frac{4\left(a+b\right)\left(b+c\right)}{b}=\frac{5\left(a+c\right)\left(b+c\right)}{c}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{3\left(a+c\right)}{a}=\frac{4\left(b+c\right)}{b}\\\frac{4\left(a+b\right)}{b}=\frac{5\left(a+c\right)}{c}\\\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3ab+3bc=4ab+4ac\\4ac+4bc=5ab+5bc\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ab-3bc+4ac=0\\5ab+bc-4ac=0\end{matrix}\right.\)
Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\) và kết hợp \(ab+bc+ca=1\) ta có hệ:
\(\left\{{}\begin{matrix}x+y+z=1\\x-3y+4z=0\\5x+y-4z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\y=\frac{1}{2}\\z=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=\frac{1}{6}\\bc=\frac{1}{2}\\ac=\frac{1}{3}\end{matrix}\right.\) (1)
\(\Rightarrow\left(abc\right)^2=\frac{1}{36}\Rightarrow abc=\pm\frac{1}{6}\) (2)
Chia vế cho vế (2) cho (1) sẽ tìm nốt được a;b;c