Giải hệ phương trình: \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=2\\x^2+y^2=2\end{matrix}\right.\)
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16 tháng 2 2020
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}2a+6b=1,1\\4a-9b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{9b+1}{4}\\\frac{2\cdot\left(9b+1\right)}{4}-9b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{-1}{9}\\a=\frac{9b+1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=\frac{-1}{9}\end{matrix}\right.\)
Pt vô nghiệm.
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 11 2019
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2y=y^2+2\\3xy^2=x^2+2\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}=\frac{y^2+2}{x^2+2}\)
\(\Rightarrow x^3+2x=y^3+2y\Rightarrow x^3-y^3+2\left(x-y\right)=0\)
\(\Rightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)
\(\Rightarrow x=y\)
Thay vào pt đầu:
\(3x^3=x^2+2\Leftrightarrow3x^3-x^2-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)
\(\Rightarrow x=y=1\)
9 tháng 3 2020
\(\left\{{}\begin{matrix}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{2}\left(xy+3x+2y+6\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{2}x+y+3=50\\x+y-2=32\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{2}x+5=18\\x+y-2=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
vậy hệ phương trình có nghiệm(x;y)=(26;8)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2019
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{x+y}{xy}=2\\ (x+y)^2-2xy=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+y=2xy\\ (x+y)^2-2xy=2\end{matrix}\right.\)
\(\Rightarrow (2xy)^2-2xy=2\)
\(\Leftrightarrow 2(xy)^2-xy-1=0\)
\(\Leftrightarrow 2xy(xy-1)+(xy-1)=0\Leftrightarrow (xy-1)(2xy+1)=0\)
\(\Leftrightarrow \left[\begin{matrix} xy=1\\ xy=\frac{-1}{2}\end{matrix}\right.\)
Nếu $xy=1\Rightarrow x+y=2xy=2$
$\Rightarrow y=2-x\Rightarrow xy=x(2-x)=1$
$\Leftrightarrow x^2-2x+1=0\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\Rightarrow y=\frac{1}{x}=1$
Nếu $xy=\frac{-1}{2}\Rightarrow x+y=2xy=-1$
$\Rightarrow y=-1-x\Rightarrow xy=x(-1-x)=\frac{-1}{2}$
$\Leftrightarrow x^2+x-\frac{1}{2}=0\Rightarrow x=\frac{-1+\sqrt{3}}{2}$
$\Rightarrow y=\frac{-1}{2x}=\frac{-1\mp \sqrt{3}}{2}$
Vậy $(x,y)=(1,1); (\frac{-1+\sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}); (\frac{-1-\sqrt{3}}{2}, \frac{-1+\sqrt{3}}{2})$
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