\(x+y=11\)
\(\dfrac{1}{10}x+\dfrac{1}{15}y=1\)
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10 tháng 2 2022
Đặt 1/x = a ; 1/y = b
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\\dfrac{10}{3}a+10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10a+10b=\dfrac{5}{3}\\\dfrac{10}{3}a+10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{20}{3}a=\dfrac{2}{3}\\b=\dfrac{1}{6}-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{10}\\b=\dfrac{1}{15}\end{matrix}\right.\)
Theo cách đặt x = 10 ; y = 15
10 tháng 2 2022
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3}.\dfrac{1}{x}+\dfrac{10}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3x}+\dfrac{10}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{3x}-\dfrac{1}{y}=\dfrac{1}{6}-\dfrac{1}{10}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}=\dfrac{1}{15}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=30\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{3.10}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{30}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{y}=\dfrac{1}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=15\end{matrix}\right.\)
2 tháng 7 2022
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-1}-\dfrac{13}{y-2}=\dfrac{2}{5}-1=-\dfrac{3}{5}\\\dfrac{-8}{x-1}+\dfrac{30}{y-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=\dfrac{-16}{5}\\y-2=-20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{11}{5}\\y=-18\end{matrix}\right.\)
T
2
10 tháng 6 2021
`1/10x+1/15(11-x)=1`
`<=>1/10x+11/15-1/15x=1`
`<=>1/30x=1-11/15=4/15`
`<=>x=4/15*30=8`
Vậy `x=8`
10 tháng 6 2021
\(\dfrac{x}{10}+\dfrac{11-x}{15}=1< =>\dfrac{3x+22-2x}{30}=1\)
\(< =>\dfrac{3x+22-2x}{30}=1=>x+22=30< =>x=30-22< =>x=8\)
XO
29 tháng 12 2021
ĐKXĐ : x;y \(\ne0\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=-2\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{1}{x}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x=\dfrac{1}{9}\end{matrix}\right.\)
TA
15 tháng 3 2023
ĐK: `x ne 2; y ne -1`
Đặt `{a=(1/(x-2)),(b=1/(y+1)):}`
Có: `{(2a+b=3),(4a-3b=1):}`
`<=>{(4a+2b=6),(4a-3b=1):}`
`<=>{(2a+b=3),(5b=5):}`
`<=>{(2a+1=3),(b=1):}`
`<=>{(a=1),(b=1):}`
``
`=>{(1/(x-2)=1),(1/(y+1)=1):}`
`<=>{(x-2=1),(y+1=1):}`
`<=>{(x=3),(y=0):}` (TM)
``
Vậy `(x;y)=(3;0)`.
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 2 2021
ĐKXĐ: ...
\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}=-\dfrac{1}{2}\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{u}{v}=-\dfrac{1}{2}\\u^2-2v=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=-2u\\u^2-2v=5\end{matrix}\right.\) \(\Rightarrow u^2+4u=5\)
\(\Leftrightarrow...\)
\(x+y=11=>y=11-x\left(1\right)\)
có: \(\dfrac{x}{10}+\dfrac{y}{15}=1\left(2\right)\)
thế(1) vào(2)=>\(\dfrac{x}{10}+\dfrac{11-x}{15}=1< =>\dfrac{3x+22-2x}{30}=1\)
\(=>x+22=30=>x=8\)(3)
thế (3) vào(1)=>y\(=11-8=3\)
\(\left\{{}\begin{matrix}x+y=11\\\dfrac{1}{10}x+\dfrac{1}{15}y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{10}x+\dfrac{1}{10}y=\dfrac{11}{10}\left(1\right)\\\dfrac{1}{10}x+\dfrac{1}{15}y=1\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\Rightarrow\dfrac{1}{30}x=\dfrac{1}{10}\Rightarrow x=3\Rightarrow y=11-3=8\)