\(\left\{{}\begin{matrix}\dfrac{3}{x-1}-\dfrac{1}{y+2}=\dfrac{3}{4}\\\dfrac{5}{x-1}+\dfrac{3}{y+2}=\dfrac{29}{12}\end{matrix}\right.\)giải chi tiết giúp mình nha☘⚽✿❤ mình cảm ơn nha
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AM
6 tháng 2 2022
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)
Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)
Lấy (4) trừ (3) ta có:
\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)
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={(\(\dfrac{17}{2};-\dfrac{17}{22}\))}
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: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
CW
17 tháng 1 2018
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
18 tháng 1 2021
b) ĐKXĐ: \(x,y\neq 0\).
Ta có: \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{1}{x}-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y=0\\xy=-1\end{matrix}\right.\\2y=x^3+1\end{matrix}\right.\).
Với x - y = 0 suy ra x = y. Do đó \(2x=x^3+1\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1=y\left(TMĐK\right)\\x=\pm\dfrac{\sqrt{5}-1}{2}=y\left(TMĐK\right)\end{matrix}\right.\).
Với xy = -1 suy ra \(y=-\dfrac{1}{x}\). Do đó \(x^3+\dfrac{2}{x}+1=0\Rightarrow x^4+x+2=0\). Phương trình vô nghiệm do \(x^4+x+2=\left(x^2-\dfrac{1}{2}\right)^2+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{2}>0\).
Vậy...
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: \(\left\{{}\begin{matrix}x< >1\\y< >1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{5x}{y-1}-\dfrac{2y}{x-1}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4x}{y-1}+\dfrac{6y}{x-1}=2\\\dfrac{15x}{y-1}-\dfrac{6y}{x-1}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{19x}{y-1}=-4\\\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{y-1}=\dfrac{-19}{4}\\2\cdot\dfrac{-19}{4}+\dfrac{3y}{x-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x=-19\left(y-1\right)\\\dfrac{3y}{x-1}=1+\dfrac{19}{2}=\dfrac{21}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\\dfrac{y}{x-1}=\dfrac{7}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+19y=19\\7x-7=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+19y=19\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8x+38y=38\\133x-38y=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}141x=171\\7x-2y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{171}{141}\\2y=7x-7=\dfrac{70}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{171}{141}=\dfrac{57}{47}\\y=\dfrac{35}{47}\end{matrix}\right.\left(nhận\right)\)
5 tháng 1 2023
a: =>2/x+2/y=2 và 4/x-2/y=1
=>6/x=3 và 1/x+1/y=1
=>x=2 và 1/y=1-1/2=1/2
=>x=2; y=2
b: Đặt 1/x=a; 1/y=b
=>1/3a+1/3b=1/4 và 5/6a+b=2/3
=>a=1/2; b=1/4
=>x=2; y=4
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