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từ phương trình số 2 ta có
\(\left(x+y\right)\left(x+2y\right)+\left(x+y\right)=0\Leftrightarrow\left(x+y\right)\left(x+2y+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\x+2y+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-y\\x=-2y-1\end{cases}}\)
lần lượt thay vào 1 ta có
\(\orbr{\begin{cases}y^2+7=y^2+4y\\\left(-2y-1\right)^2+7=y^2+4y\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{7}{4}\\3y^2+8=0\end{cases}}}\)
vậy hệ có nghiệm duy nhất \(x=-y=-\frac{7}{4}\)
Ta có:
\(\left\{{}\begin{matrix}x^2+2y^2-3xy-2x+4y=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-2x\right)-\left(2xy-4y\right)-\left(xy-2y^2\right)=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)-2y\left(x-2\right)-y\left(x-2y\right)=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x-2y\right)-y\left(x-2y\right)=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x-2y\right)=0\\x^4-10x^2+25=2x-2y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y-2=0\\x-2y=0\end{matrix}\right.\\x^4-10x^2+20-2x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\x^4-10x^2+20-2x+2\left(x-2\right)=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\x^4-10x^2+20-2x+\dfrac{2x}{2}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\x^4-10x^2+16=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\x^4-10x^2-x+20=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\\left(x^2-8\right)\left(x^2-2\right)=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left(x^2-x-5\right)\left(x^2+x-4\right)=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\\left[{}\begin{matrix}x^2=8\\x^2=2\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left[{}\begin{matrix}x^2-x-5=0\\x^2+x-4=0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\\left[{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=x-2\\\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left[{}\begin{matrix}x=\dfrac{1+\sqrt{21}}{2}\\x=\dfrac{1-\sqrt{21}}{2}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{17}}{2}\\x=\dfrac{-1-\sqrt{17}}{2}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}y=\sqrt{8}-2\\x=\sqrt{8}\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\sqrt{8}-2\\x=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=\sqrt{2}-2\\x=\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\sqrt{2}-2\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{1+\sqrt{21}}{4}\\x=\dfrac{1+\sqrt{21}}{2}\end{matrix}\right.\\\end{matrix}\right.\) (CÒN MỘT VÀI TRƯỜNG HỢP BÊN TRÊN MK KO VIẾT HẾT ĐƯỢC BẠN TỰ TÌM Y NHA)
Xét \(x^2-3xy+y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\)
Đơn giản rồi nhé
Xét phương trình đầu: \(x^2-\left(3y+2\right)x+2y^2+4y=0\)(1)
Xem x là ẩn và y là tham số:
\(\Delta=\left(3y+2\right)^2-4\left(2y^2+4y\right)=y^2-4y+4=\left(y-2\right)^2\)
Phương trình (1) có 2 nghiệm
\(x_1=\frac{\left(3y+2\right)-\left(y-2\right)}{2}=y+2\)
\(x_2=\frac{3y+2+\left(y-2\right)}{2}=2y\)
+) Với x =y+2 <=> y=x-2Thế vào phương trình (2) Ta có:
\(\left(x^2-5\right)^2=9\Leftrightarrow\orbr{\begin{cases}x^2-5=-3\\x^2-5=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=2\\x^2=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm\sqrt{2}\\x=\pm2\sqrt{2}\end{cases}}\)
thế vào tìm y
+) Với x=2y thế vào ta có: \(\left(x^2-5\right)^2=x+5\Leftrightarrow x^4-10x^2-x+20=0\)
\(\Leftrightarrow\left(x^4-9x^2+\frac{81}{4}\right)-\left(x^2+x+\frac{1}{4}\right)=0\Leftrightarrow\left(x^2-\frac{9}{4}\right)^2-\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x^2-x-5\right)\left(x^2+x-4\right)=0\)
Em làm tiếp nhé
\(x^2-\left(3y-2\right)x+2y^2-4y=0\)
\(\Delta=\left(3y-2\right)^2-4\left(2y^2-4y\right)=y^2+4y+4=\left(y+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y-2+y+2}{2}=2y\\x=\frac{3y-2-y-2}{2}=y-2\end{matrix}\right.\)
Thế xuống dưới:
\(\Rightarrow\left[{}\begin{matrix}4y^2+y^2-2y^2+2y-5=0\\\left(y-2\right)^2+y^2-y\left(y-2\right)+2y-5=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
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