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a/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=u\\\frac{1}{2y+1}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u+v=\frac{6}{5}\\3u-2v=\frac{11}{10}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\frac{1}{2}\\v=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=2\\2y+1=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}x+y=u\\\sqrt{x+1}=v\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u+v=4\\u-3v=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=1\\\sqrt{x+1}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x+1=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
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
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/ \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{2x}{3}+\frac{x}{4}-\frac{y}{6}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\\frac{11}{12}x-\frac{y}{6}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=8\\11x-2y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{28}{13}\\y=\frac{76}{13}\end{matrix}\right.\)
\(xy\ne0\)
\(\left\{{}\begin{matrix}\frac{x}{y}-\frac{y}{x}=\frac{5}{6}\\x^2-y^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{x^2-y^2}{xy}=\frac{5}{6}\\x^2-y^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}xy=6\\x^2-y^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{6}{x}\\x^2-y^2=5\end{matrix}\right.\)
\(\Rightarrow x^2-\frac{36}{x^2}=5\Leftrightarrow x^4-5x^2-36=0\) \(\Rightarrow x^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x=3\Rightarrow y=2\\x=-3\Rightarrow y=-2\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy..............................................................................
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
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
Vậy...................................................................................
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
a, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3.3-5=4\\x=\frac{33}{11}=3\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 3;4 )
b, Làm tương tự a
c, Ta có : \(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-y+2=1\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{3}{y-1-y+2}+\frac{2}{y-1+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\3+\frac{2}{2y-2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{2}{2y-2}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\2y-2=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-1=1\\y=2\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 1;2 )
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.
\(\left\{ \begin{array}{l} \dfrac{6}{{x + y}} + \dfrac{{11}}{{x - y}} = 21\\ \dfrac{6}{{x + y}} + \dfrac{5}{{x - y}} = 9 \end{array} \right.\)
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:
\(\left\{{}\begin{matrix}\frac{1}{x+y}=-\frac{1}{6}\\\frac{1}{x-y}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}\left(x+y\right)=1\\6\left(x-y\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}x-\frac{1}{6}y=1\\6x-6y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-\frac{1}{6}\left(\frac{1+6y}{6}\right)-\frac{1}{6}y=1\\x=\frac{1+6y}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\frac{37}{12}\\x=-\frac{35}{12}\end{matrix}\right.\)