Giải hệ phương trình sau:
{15/x - 7/y = 9
{4/x + 9/y = 35
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5 tháng 10 2021
\(a,\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{2}{y}=2\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{5}{y}=3\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{5}{3}\\\dfrac{2}{x}+\dfrac{9}{5}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{5}{3}\end{matrix}\right.\)
\(b,\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}-\dfrac{135}{y}=525\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{9}{y}=35\\-\dfrac{163}{y}=489\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-27=35\\y=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{31}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
5 tháng 10 2021
a: Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{2}{y}=2\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-3\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1}{3}\\\dfrac{1}{x}=1+\dfrac{1}{y}=1+\left(-3\right)=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
TK
22 tháng 8 2021
\(\left\{{}\begin{matrix}\dfrac{5}{y}-\dfrac{7}{y}=9\\\dfrac{4}{x}-\dfrac{9}{y}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-2}{y}=9\\\dfrac{4}{x}-\dfrac{9}{y}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{9}\\\dfrac{4}{x}-\dfrac{9}{-\dfrac{2}{9}}=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{9}\\\dfrac{4}{x}=-\dfrac{11}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{9}\\x=-\dfrac{8}{11}\end{matrix}\right.\)
Vậy....
14 tháng 3 2020
Đặt \(\frac{1}{x}=a,\frac{1}{y}=b\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}60a-28b=36\\60a+135b=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-163b=-489\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\4a+9.3=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=3\\4a=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=2\\\frac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là (x;y) = (\(\frac{1}{2};\frac{1}{3}\))
NT
1
ML
2 tháng 1 2016
\(\left(x+1\right)\left(y+1\right)=2;\text{ }\left(y+1\right)\left(z+1\right)=5;\text{ }\left(z+1\right)\left(x+1\right)=10\)
\(x+1=a;\text{ }y+1=b;\text{ }z+1=c\)
\(\rightarrow ab=2;\text{ }bc=5;\text{ }ca=10\Rightarrow\left(abc\right)^2=100\Rightarrow abc=\pm10\)
\(+abc=10:\text{ }c=\frac{abc}{ab}=\frac{10}{2}=5\), tương tự với a, b
\(+abc=-10\) tương tự trên.
AQ
1
12 tháng 10 2021
\(\left\{{}\begin{matrix}\left(x-5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-2x-5y+10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7y=12\\3x-y=16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x+21y=36\\3x-y=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}22y=20\\x+7y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{62}{11}\\y=\dfrac{10}{11}\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}\dfrac{15}{x}-\dfrac{7}{y}=9\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}+\dfrac{135}{y}=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-163}{y}=-489\\\dfrac{4}{x}+\dfrac{9}{y}=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)