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):}`
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`=>{(1/(x-2)=1),(1/(y+1)=1):}`
`<=>{(x-2=1),(y+1=1):}`
`<=>{(x=3),(y=0):}` (TM)
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Vậy `(x;y)=(3;0)`.
giải hpt:
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)
giải HPT
\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=-1\\\dfrac{3}{x}-\dfrac{2}{y}=5\end{matrix}\right.\)
giải hpt:
\(\left\{{}\begin{matrix}x+\dfrac{1}{x}+y-\dfrac{1}{y}=3\\x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=5\end{matrix}\right.\)
Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}y=2\sqrt{x-1}\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}xy-\dfrac{x}{y}=9.6\\xy-\dfrac{y}{x}=7.5\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\\\dfrac{2}{xy}-\dfrac{1}{z^2}=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x+y+1}{x+2y}+\dfrac{x+2y}{x+y+1}=2\\3x+y=4\end{matrix}\right.\)
giải hpt
\(\left\{{}\begin{matrix}\dfrac{5x-1}{x+3}-\dfrac{y+5}{y-2}\\\dfrac{x+4}{x+3}+\dfrac{2y-1}{y-2}\end{matrix}\right.\)
Giải các hệ PT sau bằng phương pháp cộng đại số
a)\(\left\{{}\begin{matrix}\text{3x-2y=1}\\\text{ 2x+4y=3}\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\text{4x-3y=1}\\\text{ -x+2y=1}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{4}{3}y=1\\\dfrac{1}{2}x-\dfrac{3}{4}y=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y+\dfrac{2}{x}+\dfrac{1}{y}=\dfrac{19}{3}\\xy+\dfrac{1}{xy}=\dfrac{10}{3}\end{matrix}\right.\)
