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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
ĐKXĐ : \(\left\{{}\begin{matrix}x>7\\y>-6\end{matrix}\right.\)
- Đặt \(\frac{1}{\sqrt{x-7}}=a,\frac{1}{\sqrt{y+6}}=b\) ( \(a,b\ne0\) ) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{matrix}\right.\)
( đoạn này ruễ tự giải nhoa )
=> \(\left\{{}\begin{matrix}a=\frac{1}{3}\\b=\frac{1}{6}\end{matrix}\right.\)( TM )
- Thay lại \(\frac{1}{\sqrt{x-7}}=a,\frac{1}{\sqrt{y+6}}=b\) vào hệ phương trình ta được :
\(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-7=9\\y+6=36\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=16\\y=30\end{matrix}\right.\) ( TM )
Vậy .........
\(\left\{ \begin{array}{l} \dfrac{6}{{x + y}} + \dfrac{{11}}{{x - y}} = 21\\ \dfrac{6}{{x + y}} + \dfrac{5}{{x - y}} = 9 \end{array} \right.\)
Đặt \(\left\{ \begin{array}{l} t = \dfrac{1}{{x + y}}\\ r = \dfrac{1}{{x - y}} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 6t - 11r = 21\\ 6t + 5r = 9 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} t = \dfrac{{17}}{8}\\ r = - \dfrac{3}{4} \end{array} \right.\)
Với \(\left\{ \begin{array}{l} t = \dfrac{{17}}{8}\\ r = - \dfrac{3}{4} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \dfrac{1}{{x + y}} = \dfrac{{17}}{8}\\ \dfrac{1}{{x - y}} = - \dfrac{3}{4} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = - \dfrac{{22}}{{51}}\\ y = \dfrac{{46}}{{51}} \end{array} \right.\)
\(\left\{{}\begin{matrix}\frac{6}{x+y}+\frac{11}{x-y}=21\\\frac{6}{x+y}+\frac{5}{x-y}=9\end{matrix}\right.\) (*)
Đặt \(\frac{1}{x+y}\) là a; \(\frac{1}{x-y}\) là b.
Phương trình (*) trở thành:
\(\left\{{}\begin{matrix}6a+11b=21\\6a+5b=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6b=12\\6a+5b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=-\frac{1}{6}\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}\frac{1}{x+y}=-\frac{1}{6}\\\frac{1}{x-y}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}\left(x+y\right)=1\\6\left(x-y\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}x-\frac{1}{6}y=1\\6x-6y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}\left(\frac{1+6y}{6}\right)-\frac{1}{6}y=1\\x=\frac{1+6y}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\frac{37}{12}\\x=-\frac{35}{12}\end{matrix}\right.\)
Pt (1) có: \(\left|y+\frac{1}{x}\right|+\left|\frac{13}{6}+x-y\right|\ge\left|\frac{13}{6}+\frac{1}{x}+x\right|\)
=> \(\frac{13}{6}+x+\frac{1}{x}\ge\left|\frac{13}{6}+x+\frac{1}{x}\right|\)
Dấu "=" xảy ra <=> \(\frac{13}{6}+x+\frac{1}{x}=0\)
<=> \(6x^2+13x+6=0\) <=>\(\left(3x+2\right)\left(2x+3\right)=0\)
<=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{matrix}\right.\)
Tại \(x=-\frac{2}{3}\) thay vào pt (2) => \(y^2=\frac{9}{4}\) =>\(\left[{}\begin{matrix}y=\frac{3}{2}\left(tm\right)\\y=-\frac{3}{2}\left(ktm\right)\end{matrix}\right.\)
Tại \(x=-\frac{3}{2}\) thay vào (2) => \(y^2=\frac{4}{9}\) => \(\left[{}\begin{matrix}y=\frac{2}{3}\left(ktm\right)\\y=-\frac{2}{3}\left(tm\right)\end{matrix}\right.\)
Vậy hpt có 2 ngiệm \(\left(-\frac{2}{3};\frac{3}{2}\right),\left(\frac{-3}{2},\frac{-2}{3}\right)\).
\(xy\ne0\)
\(\left\{{}\begin{matrix}\frac{x}{y}-\frac{y}{x}=\frac{5}{6}\\x^2-y^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{x^2-y^2}{xy}=\frac{5}{6}\\x^2-y^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}xy=6\\x^2-y^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{6}{x}\\x^2-y^2=5\end{matrix}\right.\)
\(\Rightarrow x^2-\frac{36}{x^2}=5\Leftrightarrow x^4-5x^2-36=0\) \(\Rightarrow x^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x=3\Rightarrow y=2\\x=-3\Rightarrow y=-2\end{matrix}\right.\)
Vì \(|x-2|\)nằm ở mẫu \(\Rightarrow\)x - 2 > 0 \(\Rightarrow\) x > 2 \(\Rightarrow\) \(|x-2|\)= x-2
\(\left\{{}\begin{matrix}\frac{2}{x-2}+\frac{1}{y}=2\\\frac{6}{x-2}-\frac{2}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{6}{x-2}+\frac{3}{y}=6\\\frac{6}{x-2}-\frac{2}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{y}=6\\\frac{6}{x-2}-\frac{2}{y}=1\end{matrix}\right.\)
a/ \(\Leftrightarrow\left\{{}\begin{matrix}3x-4y=11\\-x-10y=-15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\)
b/ \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{2x}{3}+\frac{x}{4}-\frac{y}{6}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{11}{12}x-\frac{y}{6}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\11x-2y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{28}{13}\\y=\frac{76}{13}\end{matrix}\right.\)
\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{x^2+y^2}{xy}=\frac{13}{6}\\ x+y=6\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{(x+y)^2-2xy}{xy}=\frac{13}{6}\\ x+y=6\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{6^2}{xy}-2=\frac{13}{6}\\ x+y=6\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} xy=\frac{216}{25}\\ x+y=6\end{matrix}\right.\)
\(\Rightarrow x(6-x)=\frac{216}{25}\)
\(\Leftrightarrow x^2-6x+\frac{216}{25}=0\)
\(\Leftrightarrow (x-3)^2=\frac{9}{25}\Rightarrow \left[\begin{matrix} x=\frac{-3}{5}+3=\frac{12}{5}\rightarrow y=6-x=\frac{18}{5}\\ x=\frac{3}{5}+3=\frac{18}{5}\rightarrow y=6-x=\frac{12}{5}\end{matrix}\right.\)
Vậy $(x,y)=(\frac{12}{5}, \frac{18}{5})$ và hoán vị.