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a) ĐK xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{7}{16};-\dfrac{42}{17}\))}
b) Đk xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.\)
Vậy S={(\(\dfrac{13}{6};\dfrac{13}{152}\))}
c) ĐK xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.\)
Vậy S={(\(-\dfrac{1}{7};\dfrac{1}{5}\))}
d) ĐK xác định : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
Vậy S={(0,4;-4)}
e) ĐKXĐ : x≠0;y≠0
ta có : \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.\) 'Vậy....
1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\)
2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\)
3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
Còn lại tương tự
Giải hệ sau :
Câu a :
\(\left\{{}\begin{matrix}x+y=-1\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy ...........................
Câu b :
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) . Ta có :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{3}{5}\\3a+4b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-b=-\dfrac{7}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{6}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{7}{5}\\\dfrac{1}{y}=-\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
câu a rút x ở pt 1 thay vào pt 2
-> đcược pt 4 ẩn tự giải đc
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=0\\-x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\left(1\right)\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\left(2\right)\end{matrix}\right.\)ĐK: \(x\ne1;y\ne-2\)
(1)\(\Leftrightarrow1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\Leftrightarrow\dfrac{2}{x-1}-\dfrac{6}{y+2}=3\)
Đặt \(A=\dfrac{1}{x-1};B=\dfrac{1}{y+2}\)
\(\Rightarrow\left\{{}\begin{matrix}2A-6B=3\\2A-5B=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{9}\\y=-1\end{matrix}\right.\)(TM)
Vậy hpt có nghiệm là \(\left(\dfrac{11}{9};-1\right)\).
b)ĐK: \(y\ge-1\)
Đặt \(A=x^2-2x;B=\sqrt{y+1}\left(B\ge0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}2A+B=0\\3A-2B=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}A=-1\\B=2\end{matrix}\right.\)(TM)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=1\end{matrix}\right.\)
Vậy hpt có nghiệm là (-1;1);(1;1).
\(Đặt:a=\dfrac{1}{x};b=\dfrac{1}{y}\left(x,y\ne0\right)\\ \left\{{}\begin{matrix}\dfrac{108}{x}+\dfrac{63}{y}=7\\\dfrac{81}{x}+\dfrac{84}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}108a+63b=7\\81a+84b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}324a+189b=21\\324a+336b=28\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-147b=-7\\81a+84b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{-7}{-147}=\dfrac{1}{21}\\81a+84.\dfrac{1}{21}=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{21}\\81a=7-4=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{1}{y}=\dfrac{1}{21}\left(TM\right)\\a=\dfrac{1}{x}=\dfrac{3}{81}=\dfrac{1}{27}\left(TM\right)\end{matrix}\right.\\ Vậy:\left\{{}\begin{matrix}x=27\\y=21\end{matrix}\right. \)