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a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
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b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
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c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}x+y=a\\x-y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a=5b\\\frac{20}{a}+\frac{20}{b}=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{2a}{5}\\\frac{20}{a}+\frac{20}{b}=7\end{matrix}\right.\)
\(\Rightarrow\frac{20}{a}+\frac{50}{a}=7\)
\(\Rightarrow\frac{70}{a}=7\Rightarrow a=10\Rightarrow b=4\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}\left(x^2+x\right)+\left(y^2+y\right)=18\\\left(x^2+x\right)\left(y^2+y\right)=72\end{matrix}\right.\)
Theo Viet đảo, \(x^2+x\) và \(y^2+y\) là nghiệm của:
\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\x=\frac{3y-1}{y}\end{matrix}\right.\)
Nhận thấy \(y=\frac{1}{3}\) không phải nghiệm
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)
\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)
\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)
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) = (\(\frac{1}{2};\frac{1}{3}\))
\(\left\{ \begin{array}{l} 4a + 6b = \dfrac{{23}}{{20}} (1)\\ 7a + 10b = \dfrac{{39}}{{20}}(2) \end{array} \right.\)
Từ $(1) \Rightarrow x =\dfrac{23}{80}-\dfrac{3}{2}y$.
Thay vào $(2)$ ta được: \(7\left( {\dfrac{{23}}{{80}} - \dfrac{3}{2}y} \right) + 10y = \dfrac{{39}}{{20}} \Rightarrow y = \dfrac{1}{8} \Rightarrow x = \dfrac{1}{10}\)