\(\left\{{}\begin{matrix}x+y=1000\\\dfrac{15}{100}x+\dfrac{17}{100}y=1162\end{matrix}\right.\)
K
Khách
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T
1
NA
1 tháng 3 2022
\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\left(x+y\right)+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\1500+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\dfrac{3}{4}x+\dfrac{17}{25}y=1083\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=900\\y=600\end{matrix}\right.\)
L
1
18 tháng 1
ĐKXĐ: \(x\notin\left\{0;-20\right\}\)
\(\left\{{}\begin{matrix}y=20+x\\\dfrac{100}{x}-\dfrac{100}{20+x}=\dfrac{5}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{400}{x\left(x+20\right)}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+20\\x\left(x+20\right)=2400\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+20\\x^2+20x-2400=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+20\\\left(x+60\right)\left(x-40\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+20\\\left[{}\begin{matrix}x+60=0\\x-40=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+20\\x\in\left\{-60;40\right\}\end{matrix}\right.\)(nhận)
=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-60\\y=x+20=-60+20=-40\end{matrix}\right.\\\left\{{}\begin{matrix}x=40\\y=x+20=60\end{matrix}\right.\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y\right)^2+\dfrac{1}{\left(x+y\right)^2}+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y+\dfrac{1}{x+y}\right)^2+\left(x-y+\dfrac{1}{x-y}\right)^2=\dfrac{136}{9}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=u\\x-y+\dfrac{1}{x-y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{16}{3}\\u^2+v^2=\dfrac{136}{9}\end{matrix}\right.\)
Hệ cơ bản, chắc bạn tự giải quyết phần còn lại được
3 tháng 12 2021
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
29 tháng 4 2021
Lời giải:
Phương hướng giải là bạn sử dụng phương pháp thế, biểu diễn $x$ theo $y$ qua 1 trong 2 PT, sau đó thế vô PT còn lại giải PT 1 ẩn $y$
a) \(\left\{\begin{matrix}
x-6y=17\\
5x+y=23\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x=17+6y\\
5x+y=23\end{matrix}\right.\)
\(\Rightarrow 5(17+6y)+y=23\)
\(\Leftrightarrow 31y=-62\Leftrightarrow y=-2\)
$x=17+6y=17+6(-2)=5$
Vậy $(x,y)=(5,-2)$
Các phần còn lại bạn giải tương tự
b) $(x,y)=(\frac{1}{4}, 0)$
c) $(x,y)=(3, 4)$
d) $(x,y)=(\frac{79}{21}, \frac{44}{21})$
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: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
A
14 tháng 2 2021
\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=9\\\dfrac{7}{x}+\dfrac{4}{y}=17\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}4a+3b=9\\7a+4b=17\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}16a+12b=36\\21a+12b=51\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-5a=-15\\4a+3b=9\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=3\\4.3+3b=9\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3\\\dfrac{1}{y}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
Vậy ...
14 tháng 2 2021
Ta có: \(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=9\\\dfrac{7}{x}+\dfrac{4}{y}=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{28}{x}+\dfrac{21}{y}=63\\\dfrac{28}{x}+\dfrac{16}{y}=68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-5\\\dfrac{4}{x}+\dfrac{3}{y}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{-1}=9\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-3=9\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}=12\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
9 tháng 2 2021
Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\) ta được hệ : \(\left\{{}\begin{matrix}4a+3b=9\\7a+4b=17\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)
- Thay lại ta được : \(\left\{{}\begin{matrix}\dfrac{1}{x}=3\\\dfrac{1}{y}=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)
Vậy ...
\(\left\{{}\begin{matrix}x+y=1000\\\dfrac{15}{100}x+\dfrac{17}{100}y=1162\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{15}{100}x+\dfrac{15}{100}y=150\\\dfrac{15}{100}x+\dfrac{17}{100}y=1162\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{2}{100}y=150-1162=-1012\\x+y=1000\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=50600\\x=1000-50600=-49600\end{matrix}\right.\)