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\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{2}\\\dfrac{x+5}{2}=\dfrac{x+7}{3}-4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\left(2x+1\right)}{12}-\dfrac{4\left(y-2\right)}{12}=\dfrac{6}{12}\\\dfrac{3\left(x+5\right)}{6}=\dfrac{2\left(x+7\right)}{6}-\dfrac{24}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3\left(2x+1\right)-4\left(y-2\right)=6\\3\left(x+5\right)=2\left(x+7\right)-24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+3-4y+8=6\\3x+15=2y+14-24\end{matrix}\right.\\ \Leftrightarrow\Leftrightarrow\left\{{}\begin{matrix}6x-4y+11=6\\3x+15=2y-10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-4y=-5\\3x-2y=-25\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2\left(3x-2y\right)=-5\\3x-2y=-25\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x-2y=-\dfrac{5}{2}\\3x-2y=-25\left(vô.lí\right)\end{matrix}\right.\)
Vậy hệ phương trình vô nghiệm
6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y
(Các câu khác tương tự nhé.)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3.4}{x+y}+\dfrac{3.4}{x-y}=\dfrac{5}{2}\\\dfrac{4}{x+y}+\dfrac{2.4}{x+y}=\dfrac{4}{3}\end{matrix}\right.\\ Đặt.a=\dfrac{4}{x+y},b=\dfrac{4}{x-y}\\ \Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{5}{2}\\a+2b=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{5}{2}\\3a+6b=4\end{matrix}\right.\)
\(\Leftrightarrow\left(3a+6b\right)-3a-3b=4-\dfrac{5}{2}\\ \Leftrightarrow3b=\dfrac{3}{2}\Rightarrow b=\dfrac{1}{2}\Rightarrow a+2.\dfrac{1}{2}=\dfrac{4}{3}\\\Leftrightarrow a+1=\dfrac{4}{3}\Rightarrow a=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{4}{x+y}=\dfrac{1}{3}\\\dfrac{4}{x+y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=12\\x-y=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\y=2\end{matrix}\right.\)
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{7}{\sqrt{x}-7}-\dfrac{4}{\sqrt{y}+6}=\dfrac{5}{3}.\\\dfrac{5}{\sqrt{x}-7}+\dfrac{3}{\sqrt{y}+6}=2\dfrac{1}{6}.\end{matrix}\right.\) \(\left(x,y\ge0;x\ne49\right).\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\dfrac{1}{\sqrt{x}-7}-4\dfrac{1}{\sqrt{y}+6}=\dfrac{5}{3}.\\5\dfrac{1}{\sqrt{x}-7}+3\dfrac{1}{\sqrt{y}+6}=\dfrac{13}{6}.\end{matrix}\right.\)
Đặt \(\dfrac{1}{\sqrt[]{x}-7}=a\); \(\dfrac{1}{\sqrt[]{y}+6}=b\left(a,b\ne0\right).\)
\(\Rightarrow\left\{{}\begin{matrix}7a-4b=\dfrac{5}{3}.\\5a+3b=\dfrac{13}{6}.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}.\\b=\dfrac{1}{6}.\end{matrix}\right.\) \(\left(TM\right).\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}-7}=\dfrac{1}{3}.\\\dfrac{1}{\sqrt{y}+6}=\dfrac{1}{6}.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-7=3.\\\sqrt{y}+6=6.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=10.\\\sqrt{y}=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=100\left(TM\right).\\y=0\left(TM\right).\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là: \(\left(x;y\right)=\left(100;0\right).\)
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{1}{4}-\dfrac{y}{3}+\dfrac{2}{3}=\dfrac{1}{12}\\\dfrac{x}{2}+\dfrac{5}{2}-\dfrac{y}{3}-\dfrac{7}{3}=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{5}{6}\\\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{25}{6}\end{matrix}\right.\) (vô lý)
Vậy HPT vô nghiệm