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1. \(\left\{{}\begin{matrix}3x^2+y^2+4xy=8\left(1\right)\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)
=> \(3x^2+3xy+xy+y^2=\left(x+y\right)\left(x^2+xy+2\right)\)
<=> \(\left(x+y\right)\left(3x+y\right)=\left(x+y\right)\left(x^2+xy+2\right)=0\)
<=> \(\left(x+y\right)\left(x^2+xy+2-3x-y\right)=0\)
<=> \(\left[{}\begin{matrix}x=-y\\x^2+xy+2-3x-y=0\end{matrix}\right.\)
TH1: x = -y thay vào pt (1), ta được:
3y2 + y2 - 4y2 = 8
<=> 0y = 8 (vô lí)
TH2: \(x^2+xy+2-3x-y=0\)
<=> x (x + y) - (x + y) - 2(x - 1) = 0
<=> (x - 1)(x + y) - 2(X - 1) = 0
<=> (x - 1)(x + y - 2) = 0
<=> \(\left[{}\begin{matrix}x=1\\x+y-2=0\end{matrix}\right.\)
Với x = 1 thay vào pt (1) -> 3 + y2 + 4y = 8
<=> y2 + 4y - 5 = 0 <=> (y + 5)(y - 1) = 0
<=> \(\left[{}\begin{matrix}y=-5\\y=1\end{matrix}\right.\)
Với x + y - 2 = 0 => x = 2 - y thay vào pt (1)
=> 3(2 - y)2 + y2 + 4(2 - y)y = 8
<=> 3y2 - 12y + 12 + y2 + 8 - 4y2 = 8
<=> 12 = 12y <=> y= 1 => x = 2 - 1 = 1
Vậy ....
Bạn coi lại đề, hệ này ko giải được
Pt bên dưới là \(xy\left(y^2+3y+3\right)=4\) thì giải được
ĐKXĐ: ...
\(xy^2+4y^2+8=x^2+2x\)
\(\Leftrightarrow x^2-\left(y^2-2\right)x-4y^2-8=0\)
\(\Delta=\left(y^2-2\right)^2+4\left(4y^2+8\right)=\left(y^2+6\right)^2\)
Phương trình có 2 nghiệm: \(\left[{}\begin{matrix}x=\frac{y^2-2-\left(y^2+6\right)}{2}=-4\\x=\frac{y^2-2+y^2+6}{2}=y^2+2\end{matrix}\right.\)
- Với \(x=-4\) thay xuống dưới:
\(y-1=3\sqrt{2y-1}\) (\(y\ge1\))
\(\Leftrightarrow y^2-2y+1=9\left(2y-1\right)\)
\(\Leftrightarrow...\)
- Với \(x=y^2+2\) thay xuống dưới:
\(y^2+y+5=3\sqrt{2y-1}\)
\(\Leftrightarrow y^2-y+\frac{1}{4}+\left(2y-1-3\sqrt{2y-1}+\frac{9}{4}\right)+\frac{7}{2}=0\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right)^2+\left(\sqrt{2y-1}-\frac{3}{2}\right)^2+\frac{7}{2}=0\) (vô nghiệm)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)
a. \(\left\{\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\left(y-2\right)\left(x-1\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\left[\begin{matrix}y-2=0\\x-1=0\end{matrix}\right.\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}\left[\begin{matrix}y=2\\x=1\end{matrix}\right.\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\left\{\begin{matrix}y=2\\3x+y=8\end{matrix}\right.\\\left\{\begin{matrix}x=1\\3x+y=8\end{matrix}\right.\end{matrix}\right.\)
Giải hệ phương trình ta được:
\(\left[\begin{matrix}\left\{\begin{matrix}y=2\\x=2\end{matrix}\right.\\\left\{\begin{matrix}x=1\\y=5\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ phương trình đã cho có tập nghiệm \(S=\left\{\left(2;2\right),\left(1;5\right)\right\}\)
b)\(\text{HPT}\Leftrightarrow \)\(\left\{\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)=12\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}a^2-4a=12\\b^2-2b=3\end{matrix}\right.\)\(\left(\left\{\begin{matrix}a=x+y\\b=x-y\end{matrix}\right.\right)\)
\(\Leftrightarrow\left\{\begin{matrix}\left[\begin{matrix}a=-2\\a=6\end{matrix}\right.\\\left[\begin{matrix}b=3\\b=-1\end{matrix}\right.\end{matrix}\right.\) Thay vào ...
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(3x+y\right)=8\\\left(x+y\right)\left(x^2+xy+2\right)=8\end{matrix}\right.\)
Chia vế cho vế: \(\frac{x^2+xy+2}{3x+y}=1\Leftrightarrow x^2+xy+2=3x+y\)
\(\Leftrightarrow x^2+\left(y-3\right)x-y+2=0\)
\(\Delta=\left(y-3\right)^2-4\left(-y+2\right)=\left(y+1\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{3-y+y+1}{2}=1\\x=\frac{3-y-y-1}{2}=-y+1\end{matrix}\right.\)
- Với \(x=1\Rightarrow y^2+4y-5=0\Rightarrow y=...\)
- Với \(x=-y+1\Rightarrow3\left(-y+1\right)^2+y^2+4y\left(-y+1\right)-8=0\Rightarrow y=...\)