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\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)
$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$
`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$
`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$
Vậy HPT vô nghiệm
a, \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-y\right)\left(x^2+y^2\right)=26\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)
Trừ vế theo vế \(pt\left(1\right)\) cho \(pt\left(2\right)\) ta được:
\(\Leftrightarrow\left(x-y\right)\left(x^2+y^2-2xy\right)=1\)
\(\Leftrightarrow\left(x-y\right)^3=1\)
\(\Leftrightarrow x-y=1\)
Khi đó hệ trở thành:
\(\left\{{}\begin{matrix}x^2+y^2=13\\\left(x+y\right)^2=25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\13+2xy=25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\2xy=12\end{matrix}\right.\)
Cộng vế theo vế 2 phương trình:
\(\left(x+y\right)^2=25\)
\(\Leftrightarrow x+y=\pm5\)
TH1: \(x+y=5\)
Ta có hệ: \(\left\{{}\begin{matrix}x-y=1\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
TH2: \(x+y=-5\)
Ta có hệ: \(\left\{{}\begin{matrix}x-y=1\\x+y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}2x^2+x-\dfrac{1}{y}=2\\y-y^2x-2y^2=-2\end{matrix}\right.\)
ĐK: \(y\ne0\)
\(\left\{{}\begin{matrix}2x^2+x-\dfrac{1}{y}=2\\y-y^2x-2y^2=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+x-\dfrac{1}{y}=2\\\dfrac{1}{y}-x-2=-\dfrac{2}{y^2}\end{matrix}\right.\)
Đặt \(\dfrac{1}{y}=t\), hệ trở thành:
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+x-t=2\\2t^2+t-x=2\end{matrix}\right.\)
\(\Rightarrow\left(x-t\right)\left(x+t+1\right)=0\)
\(\Leftrightarrow...\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
a: =>xy-2x+2y-4=xy+y và 5xy+10x+y+2=5xy-10x-2y+4
=>-2x+y=4 và 20x+3y=2
=>x=-5/13; y=42/13
b: =>4x+2|y|=8 và 4x-3y=1
=>2|y|-3y=7 và 4x-3y=1
TH1: y>=0
=>2y-3y=7 và 4x-3y=1
=>-y=7 và 4x-3y=1
=>y=-7(loại)
TH2: y<0
=>-2y-3y=7 và 4x-3y=1
=>y=-7/5; 4x=1+3y=1-21/5=-16/5
=>x=-4/5; y=-7/5
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)
