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\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+x=a\\y^2+y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=8\\ab=12\end{matrix}\right.\) theo Viet đảo, a và b là nghiệm:
\(t^2-8t+12=0\Rightarrow\left[{}\begin{matrix}t=6\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=2\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x-6=0\\y^2+y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x-2=0\\y^2+y-6=0\end{matrix}\right.\end{matrix}\right.\)
Bạn tự bấm máy
b/
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy+1=0\\\left(x+y\right)^2-2xy-x-y=22\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+2xy+2=0\\\left(x+y\right)^2-2xy-x-y-22=0\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+\left(x+y\right)-20=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=4\Rightarrow xy=-5\\x+y=-5\Rightarrow xy=4\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=-5\end{matrix}\right.\) thì x; y là nghiệm:
\(t^2-4t-5=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=5\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(-1;5\right);\left(5;-1\right)\)
TH2: \(\left\{{}\begin{matrix}x+y=-5\\xy=4\end{matrix}\right.\) thì x; y là nghiệm:
\(t^2+5t+4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(-1;-4\right);\left(-4;-1\right)\)
Biến đổi pt dưới:
\(x^2-4x+4+y\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+y\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2+y\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2-y\end{matrix}\right.\)
Thay vào pt đầu giải bt
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=xy+3y-1\\\left(x+y\right)\left(x^2+1\right)=x^2+y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^2+\left(x-3\right)y+x^2+1=0\\x^3+x+x^2y-x^2-1=0\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow y^2-\left(x^2-x+3\right)y-x^3+2x^2-x+2=0\)
\(\Delta=\left(x^2-x+3\right)^2-4\left(-x^3+2x^2-x+2\right)=\left(x^2+x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{x^2-x+3+x^2+x-1}{2}=x^2+1\\y=\dfrac{x^2-x+3-x^2-x+1}{2}=-x+2\end{matrix}\right.\)
Thế vào pt dưới:
\(\left[{}\begin{matrix}x+x^2+1=2\\x-x+2=\dfrac{x^2+1-x+2}{x^2+1}\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(\left\{{}\begin{matrix}1=x^2+\left(y+1\right)^2-x\left(y+1\right)\\2x^3=x+y+1\end{matrix}\right.\)
Nhân vế:
\(\Rightarrow2x^3=\left(x+y+1\right)\left[x^2+\left(y+1\right)^2-x\left(y+1\right)\right]\)
\(\Rightarrow2x^3=x^3+\left(y+1\right)^3\)
\(\Rightarrow x^3=\left(y+1\right)^3\)
\(\Rightarrow x=y+1\)
Thế vào pt đầu sẽ được 1 pt bậc 2 một ẩn
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) . Khi đó hệ phương trình trở thành :
\(\left\{{}\begin{matrix}a+b=-1\\a^2-a-2b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-a-1\left(1\right)\\a^2+a-20=0\left(2\right)\end{matrix}\right.\)
Xét phương trình (2) : \(a^2+a-20=0\)
\(\Delta=1+80=81>0\)
\(\Rightarrow\left\{{}\begin{matrix}a_1=\frac{-1+9}{2}=4\\a_1=\frac{-1-9}{2}=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b_1=-5\\b_2=4\end{matrix}\right.\)
Với \(\left(a_1;b_1\right)=\left(4;-5\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=-5\end{matrix}\right.\Rightarrow x^2-4x-5=0\)
\(\Delta=16+20=36>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{4+6}{2}=5\\x_2=\frac{4-6}{2}=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y_1=-1\\y_2=5\end{matrix}\right.\)
Với \(\left(a_2;b_2\right)=\left(-5;4\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-5\\xy=4\end{matrix}\right.\Rightarrow x^2+5x+4=0\)
\(\Delta=25-16=9>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_3=\frac{-5+3}{2}=-1\\x_4=\frac{-5-3}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y_3=-4\\y_4=-1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}\left(x_1;y_1\right)=\left(5;-1\right)\\\left(x_1;y_2\right)=\left(-1;5\right)\\\left(x_3;y_3\right)=\left(-1;-4\right)\\\left(x_4;y_4\right)=\left(-4;-1\right)\end{matrix}\right.\)
\(x+y+xy+1=0\)
\(\Leftrightarrow x\left(y+1\right)+y+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\)
Thế xuống pt dưới...