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\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (1)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = \(\dfrac{1}{x}\) ; b = \(\dfrac{1}{y}\)
Thay vào (1) ta được :
\(\left\{{}\begin{matrix}4a+b=12\\a+b=-3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}3a=15\\a+b=-3\end{matrix}\right.< =>\left\{{}\begin{matrix}a=5\\b=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{8}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{1}{5};-\dfrac{1}{8}\))}
\(b.\left\{{}\begin{matrix}5\dfrac{1}{x}+2\dfrac{1}{y}=6\\2\dfrac{1}{x}-\dfrac{1}{y}=3\end{matrix}\right.\) (2)
ĐK xác định : x≠0 ; y≠0
Đặt ẩn phụ : a = 1/x ; b = 1/y
Thay vào (2) ta được : \(\left\{{}\begin{matrix}5a+2b=6\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}5a+2b=6\\4a-2b=6\end{matrix}\right.< =>\left\{{}\begin{matrix}9a=12\\2a-b=3\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{4}{3}\\b=-\dfrac{1}{3}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-3\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{4};-3\) )}
c) \(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.\)
ĐK xác định : x≠0 ; y ≠0
Áp dụng quy tác cộng đại số ta có :
\(\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}3\dfrac{1}{x}-6\dfrac{1}{y}=2\\3\dfrac{1}{x}-3\dfrac{1}{y}=15\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-3\dfrac{1}{y}=-13\\\dfrac{1}{x}-\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{3}{13}\\x=\dfrac{3}{28}\end{matrix}\right.\)
Vậy S = {(\(\dfrac{3}{28};\dfrac{3}{13}\))}
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\)
ĐK xác định : x≠0 ; y≠0
áp dụng quy tắc cộng đại số ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{x}-4\dfrac{1}{y}=5\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.< =>\left\{{}\begin{matrix}2\dfrac{1}{x}-8\dfrac{1}{y}=10\\2\dfrac{1}{x}-3\dfrac{1}{y}=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-5\dfrac{1}{y}=9\\\dfrac{1}{x}-4\dfrac{1}{y}=5\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{5}{9}\\x=-\dfrac{5}{11}\end{matrix}\right.\)
Vậy S = {(\(-\dfrac{5}{11};-\dfrac{5}{9}\))}
e) ĐK xác định x≠0 ; y≠0
\(\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\6\dfrac{1}{x}-\dfrac{1}{y}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{1}{x}-3\dfrac{1}{y}=4\\18\dfrac{1}{x}-3\dfrac{1}{y}=6\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-17\dfrac{1}{x}=-2\\\dfrac{1}{x}-3\dfrac{1}{y}=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=\dfrac{17}{2}\\y=-\dfrac{17}{22}\end{matrix}\right.\)
Vậy S={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+y}-\dfrac{4}{x-y}=0\\\dfrac{40}{x+y}+\dfrac{40}{x-y}=9\end{matrix}\right.\)
Đặt \(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b\) . Khi đó hệ thành :
\(\left\{{}\begin{matrix}5a-4b=0\\40a+40b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}40a-32b=0\\40a+40b=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-72b=-9\\40a-32b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{10}\\b=\dfrac{1}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\x-y=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(9;1\right)\)
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)
Đặt \(\dfrac{x}{y}=a;\dfrac{x}{y+12}=b;\dfrac{x}{y-12}=c\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\c-a=2\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{2}{a}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(\dfrac{1}{b}-\dfrac{1}{a}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
\(\Leftrightarrow\dfrac{a-b}{ab}+\dfrac{a-c}{ac}=0\Leftrightarrow\dfrac{1}{ab}-\dfrac{2}{ac}=0\)
\(\Leftrightarrow\dfrac{1}{a}\left(\dfrac{1}{b}-\dfrac{2}{c}\right)=0\Rightarrow\dfrac{1}{b}=\dfrac{2}{c}\Rightarrow c=2b\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\2b-a=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=4\\\dfrac{x}{y+12}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4y\\\dfrac{4y}{y+12}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=144\\y=36\end{matrix}\right.\)
Vậy . . .
ĐK: y\(\ne0,y\ne\pm12\)
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y-12}-\dfrac{x}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\dfrac{xy+12x-xy}{y\left(y+12\right)}=1\\\dfrac{xy-xy+12x}{y\left(y-12\right)}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}12x=y^2+12y\\12x=2y^2-24y\end{matrix}\right.\)\(\Leftrightarrow\)\(y^2+12y=2y^2-24y\Leftrightarrow y^2-36y=0\Leftrightarrow y\left(y-36\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y=0\left(ktm\right)\\y=36\left(tm\right)\end{matrix}\right.\)\(\Leftrightarrow y=36\Leftrightarrow12x=36^2+12.36\Leftrightarrow x=144\)
Vậy (x;y)=(144;36)
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
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....
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