\(\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{3}{y+2}=\dfrac{1}{2}\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=\dfrac{1}{18}\end{matrix}\right.\) Làm chi tiết giúp ạ!
K
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23 tháng 1 2022
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{y+2}=\dfrac{4}{9}\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\x-1=12\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(13;7\right)\)
8 tháng 1 2018
\(a.\left\{{}\begin{matrix}4\dfrac{1}{x}+\dfrac{1}{y}=12\\\dfrac{1}{x}+\dfrac{1}{y}=-3\end{matrix}\right.\) (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={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
CW
17 tháng 1 2018
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 1
ĐKXĐ: \(x\ne2y;x\ne\dfrac{y}{2}\)
\(\left\{{}\begin{matrix}\dfrac{2}{2x-y}+\dfrac{3}{x-2y}=\dfrac{1}{2}\\\dfrac{2}{2x-y}-\dfrac{1}{x-2y}=\dfrac{1}{18}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{4}{9}\\\dfrac{2}{2x-y}-\dfrac{1}{x-2y}=\dfrac{1}{18}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2y}=\dfrac{1}{9}\\\dfrac{1}{2x-y}=\dfrac{1}{2\left(x-2y\right)}+\dfrac{1}{36}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2y}=\dfrac{1}{9}\\\dfrac{1}{2x-y}=\dfrac{1}{12}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-2y=9\\2x-y=12\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=-2\end{matrix}\right.\)
26 tháng 11 2023
a: ĐKXĐ: x<>-1 và y<>-1
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x+2-2}{x+1}+\dfrac{y+1-1}{y+1}=2\\\dfrac{x+1-1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2-\dfrac{2}{x+1}+1-\dfrac{1}{y+1}=2\\1-\dfrac{1}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=2-3=-1\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=1+1=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{-2}{x+1}+\dfrac{-1}{y+1}=-1\\\dfrac{2}{x+1}-\dfrac{6}{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y-1}=3\\\dfrac{1}{x+1}-\dfrac{3}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-1=-\dfrac{7}{3}\\\dfrac{1}{x+1}-3:\dfrac{-7}{3}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}+3\cdot\dfrac{3}{7}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\\dfrac{1}{x+1}=2-\dfrac{9}{7}=\dfrac{5}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x+1=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{4}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: y<>0 và y<>-12
\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y}-\dfrac{x}{y+12}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\cdot\dfrac{x}{y+12}=3\left(vôlý\right)\\\dfrac{x}{y}-\dfrac{x}{y+12}=1\end{matrix}\right.\)
Vậy: \(\left(x,y\right)\in\varnothing\)
d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >1\\y< >1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{5x}{y-1}-\dfrac{2y}{x-1}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4x}{y-1}+\dfrac{6y}{x-1}=2\\\dfrac{15x}{y-1}-\dfrac{6y}{x-1}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19x}{y-1}=-4\\\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y-1}=\dfrac{-19}{4}\\2\cdot\dfrac{-19}{4}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x=-19\left(y-1\right)\\\dfrac{3y}{x-1}=1+\dfrac{19}{2}=\dfrac{21}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\\dfrac{y}{x-1}=\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+19y=19\\7x-7=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+38y=38\\133x-38y=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}141x=171\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{171}{141}\\2y=7x-7=\dfrac{70}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{171}{141}=\dfrac{57}{47}\\y=\dfrac{35}{47}\end{matrix}\right.\left(nhận\right)\)
3 tháng 12 2021
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
14 tháng 6 2022
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{y+2}+\dfrac{1}{y+2}=\dfrac{1}{2}-\dfrac{1}{18}\\\dfrac{2}{x-1}-\dfrac{1}{y+2}=\dfrac{1}{18}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\\dfrac{2}{x-1}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=9\\x-1=12\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(13;7\right)\)