ĐKXĐ: \(xy\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{8}{x}+\frac{6}{y}=-2\\\frac{9}{x}-\frac{6}{y}=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{8}{x}+\frac{6}{y}=-2\\\frac{17}{x}=13\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{17}{13}\\y=-\frac{17}{23}\end{matrix}\right.\)
1.Giải hệ phương trình:
a.\(\left\{{}\begin{matrix}2\sqrt{2}x+y=2\sqrt{2}\\7x-3y=7\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}7x+y=-\frac{1}{7}\\-\frac{4}{3}x-2y=1\frac{1}{3}\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2\sqrt{5}x+3y=\sqrt{2}\\\sqrt{5}x-y=3\sqrt{2}\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y}=-5\\\frac{3}{x}-\frac{4}{y}=1\end{matrix}\right.\)
e.\(\left\{{}\begin{matrix}-\frac{5}{3x+1}+\frac{7}{2x+1}=\frac{5}{7}\\\frac{1}{3x+1}-\frac{1}{2y-3}=\frac{2}{7}\\\end{matrix}\right.\)
g.\(\left\{{}\begin{matrix}2x^2+5y^2=129\\-3x^2+y^2=13\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}2x+4=0\\4x+2y=-3\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}\left(x-15\right).\left(y+2\right)=x.y\\\left(x+15\right).\left(y-1\right)=x.y\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}2x+4=y\\x+2y=-3\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=5\\\frac{2}{x}+\frac{5}{y}=7\end{matrix}\right.\) tính bằng phương pháp cộng dại số
1)\(\left\{{}\begin{matrix}x^2-y^2-2x+2y=0\\x^2-3xy+5y^2-3=0\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{1-y}=1\\\frac{1}{x-1}-\frac{1}{y}=2\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x^2-4x+3=0\\x^2+xy+y^2=1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2+y^2+x+y=2\\\left(x+1\right)^2-\left(y+2\right)^2=0\end{matrix}\right.\)
bài 2: Tính hai cạnh góc vuông của một tâm giác vuông có độ dài cạnh huyền = 37m và diện tích = 210\(m^2\)
bài 3: giải hệ pt sau:
a. \(\left\{{}\begin{matrix}\frac{x-y}{7}+\frac{2x+y}{17}=7\\\frac{4x+y}{5}+\frac{y-7}{19}=15\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}5\left(x+2y\right)-3\left(x-y\right)=99\\x-y=7\left(x-y\right)+3y=17\end{matrix}\right.\)
c. \(\frac{x}{x-1}-\frac{2\sqrt{2}}{1-x}-\frac{6+\sqrt{2}}{x^2-1}\)= 0
d. \(\left\{{}\begin{matrix}\frac{2}{x+3}-\frac{5}{y-2}=1\\\frac{x+4}{x+3}+\frac{y}{y-2}=2\end{matrix}\right.\)
e. \(\left\{{}\begin{matrix}\frac{x-y}{7}+\frac{2x+y}{17}=7\\\frac{4+y}{5}+\frac{y-7}{19}=15\end{matrix}\right.\)
Bµi 1: A)\(\left\{{}\begin{matrix}X=35.\left(Y+2\right)\\X=50.\left(Y-1\right)\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}Y=2X-3\\Y=X-1\end{matrix}\right.\)
C) \(\left\{{}\begin{matrix}\left(X+14\right).\left(Y-2\right)=X.Y\\\left(X-4\right).\left(Y+1\right)=X.Y\end{matrix}\right.\)
D)\(\left\{{}\begin{matrix}Y=\frac{6-X}{4}\\Y=\frac{4X-5}{3}\end{matrix}\right.\)GIẢI BÀI 1 BẰNG PHƯƠNG PHAP THẾ
\(\left\{{}\begin{matrix}\frac{2}{x-1}+\frac{3}{y-2}=\frac{11}{3}\\\frac{3}{x-1}-\frac{4}{y-2}=-3\end{matrix}\right.\)
Giải phương trình bằng phương pháp thế
\(\left\{{}\begin{matrix}\frac{2x-3y}{4}-\frac{x+y-1}{5}=2x-y-1\\\frac{4x+y-2}{4}=\frac{2x-y-3}{6}-\frac{x-y-1}{3}\end{matrix}\right.\)
Giải các hệ phương trình sau bằng phương pháp thế:
a\(\left\{{}\begin{matrix}x+3y=4\\4x=5y=18\end{matrix}\right.\)
b\(\left\{{}\begin{matrix}5x-3y=5\\2x+5y=33\end{matrix}\right.\)
c\(\left\{{}\begin{matrix}\frac{x}{2}-\frac{y}{3}=0\\5x+y=13\end{matrix}\right.\)
d\(\left\{{}\begin{matrix}x+2y=\frac{7}{6}\\4x+6y=4\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}\frac{3}{2x-y}-x=2-\frac{2x^2+xy+5}{2x+y}\\\frac{1}{2x-y}-\frac{1}{2x+y}=\frac{2}{15}\end{matrix}\right.\)
