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ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{2}{x-y}+\frac{6}{x+y}=1,1\\\frac{4}{x-y}-\frac{9}{x+y}=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{4}{x-y}+\frac{12}{x+y}=2,2\\\frac{4}{x-y}-\frac{9}{x+y}=0,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\frac{21}{x+y}=2,1\\\frac{2}{x-y}=1,1-\frac{6}{x+y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\\frac{2}{x-y}=1,1-\frac{6}{x+y}=1,1-\frac{6}{10}=\frac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)(tm)
Vậy hệ phương trình có nghiệm duy nhất là (x;y) = (7;3)
another way to solve
Đặt \(\left\{{}\begin{matrix}\frac{1}{x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\)
\(hpt\Leftrightarrow\left\{{}\begin{matrix}2a+6b=1,1\\4a-9b=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1,1-6b}{2}\\\frac{4\cdot\left(1,1-6b\right)}{2}-9b=0,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{1}{10}\\a=\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x-y}=\frac{1}{4}\\\frac{1}{x+y}=\frac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x+y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)
Vậy....
\(\frac{x^2}{\left(y+1\right)^2}+\frac{y^2}{\left(x+1\right)^2}=\frac{1}{2}\Leftrightarrow\left(\frac{x}{y+1}+\frac{y}{x+1}\right)^2=\frac{1}{2}+\frac{2xy}{xy+x+y+1}\)
\(\Leftrightarrow\left(\frac{x^2+x+y^2+y}{xy+x+y+1}\right)^2=\frac{1}{2}+\frac{2xy}{4xy}\)
\(\Leftrightarrow\left(\frac{\left(x+y\right)^2-2xy+\left(x+y\right)}{4xy}\right)^2=1\)
\(\Leftrightarrow\left(\frac{\left(3xy-1\right)^2+xy-1}{4xy}\right)^2=1\)
Đặt s=x+y;p=xy (s2\(\ge\)4p)
Suy ra: \(\left(\frac{\left(3p-1\right)^2+p-1}{4p}\right)^2=1\)
=>\(\frac{9p^2-5p}{4p}=1\)hoặc \(\frac{9p^2-5p}{4p}=-1\)
<=>p=1 hoặc p=1/9
Với p=1 thì: 3=s+1=>s=2 (thỏa dk)
=>nghiệm của hpt là nghiệm của pt: X2-2X+1=0
=>x=1
Vậy hpt có 1 nghiệm là: (1;1)
Với p=1/9=>s=-2/3 (thỏa dk)
Giải như trên òi kết luận
Ta có : \(\left\{{}\begin{matrix}\frac{2x-3y}{4}-\frac{x+y-1}{5}=2x-y-1\\\frac{4x+y-2}{4}=\frac{2x-y-3}{6}-\frac{x-y-1}{3}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{5\left(2x-3y\right)}{20}-\frac{4\left(x+y-1\right)}{20}=\frac{20\left(2x-y-1\right)}{20}\\\frac{3\left(4x+y-2\right)}{12}=\frac{2\left(2x-y-3\right)}{12}-\frac{4\left(x-y-1\right)}{12}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5\left(2x-3y\right)-4\left(x+y-1\right)=20\left(2x-y-1\right)\\3\left(4x+y-2\right)=2\left(2x-y-3\right)-4\left(x-y-1\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\12x+3y-6=4x-2y-6-4x+4y+4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-15y-4x-4y+4-40x+20y+20=0\\12x+3y-6-4x+2y+6+4x-4y-4=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-34x+y=-24\\12x+y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-24+34x\\12x-24+34x=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-24+34x\\46x=28\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=-\frac{76}{23}\\x=\frac{14}{23}\end{matrix}\right.\)
Vậy hệ phương trình trên có nghiệm là ( x;y ) = \(\left(\frac{14}{23};-\frac{76}{23}\right)\)
\(\int^{3y-2x=1}_{7y-5x=1}\Leftrightarrow\int^{3y-2x=1}_{7y-5x=3y-2x}\Leftrightarrow\int^{3y-2x=1}_{4y=3x}\Leftrightarrow\int^{\frac{9}{4}x-2x=1}_{y=\frac{3}{4}x}\Leftrightarrow\int^{x=4}_{y=3}\)