Giải hệ phương trình: \(\left\{{}\begin{matrix}x^2+2y^2-3xy-2x+4y=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)
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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2020
\(x^2-\left(3y-2\right)x+2y^2-4y=0\)
\(\Delta=\left(3y-2\right)^2-4\left(2y^2-4y\right)=y^2+4y+4=\left(y+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y-2+y+2}{2}=2y\\x=\frac{3y-2-y-2}{2}=y-2\end{matrix}\right.\)
Thế xuống dưới:
\(\Rightarrow\left[{}\begin{matrix}4y^2+y^2-2y^2+2y-5=0\\\left(y-2\right)^2+y^2-y\left(y-2\right)+2y-5=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
LL
1 tháng 5 2021
a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;0\right\}\)
LL
1 tháng 5 2021
b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;1\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 5 2019
\(x^2-\left(3y+2\right)x+2y^2+4y=0\)
\(\Delta=\left(3y+2\right)^2-4\left(2y^2+4y\right)=y^2-4y+4=\left(y-2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y+2-y+2}{2}=y+2\\x=\frac{3y+2+y-2}{2}=2y\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x-2\\2y=x\end{matrix}\right.\)
TH1: \(\) \(y=x-2\)
\(\left(x^2-5\right)^2=2x-2\left(x-2\right)+5\)
\(\Leftrightarrow\left(x^2-5\right)^2=9\Rightarrow\left[{}\begin{matrix}x^2-5=3\\x^2-5=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=8\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\sqrt{2}\Rightarrow y=-2\pm2\sqrt{2}\\x=\pm\sqrt{2}\Rightarrow y=-2\pm\sqrt{2}\end{matrix}\right.\)
TH2: \(2y=x\)
\(\Leftrightarrow\left(x^2-5\right)^2=2x-x+5\Leftrightarrow\left(x^2-5\right)^2=x+5\)
Đặt \(x^2-5=a\Rightarrow5=x^2-a\) pt trở thành:
\(a^2=x+x^2-a\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a\right)+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-x=0\\a+x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-5=0\\x^2-5+x+1=0\end{matrix}\right.\) \(\Leftrightarrow...\)
Bạnt ự giải nốt
8 tháng 1 2021
1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)
=> Hệ có vô số nghiệm.
3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
8 tháng 1 2021
1)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;2\right)\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
3)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
4)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
XO
18 tháng 3 2023
1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=0\\4x+2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
12 tháng 9 2018
mấy bài dạng như này mk sẽ hướng dẩn nha .
a) ta có : \(\left\{{}\begin{matrix}\left(x+y-2\right)\left(2x-y\right)=0\\x^2+y^2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+y-2=0\\2x-y=0\end{matrix}\right.\\x^2+y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\x^2+y^2=2\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y=0\\x^2+y^2=0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\) giải bằng cách thế bình thường nha
b) ta có : \(\left\{{}\begin{matrix}x^2+y^2+2x+2y=6\\x+y-3xy+1=0\end{matrix}\right.\) \(\Leftrightarrow2x^2+2y^2+6xy-5=0\)
\(\Leftrightarrow2\left(x+y\right)^2+2xy-5=0\) sài vi ét --> .......................
c) đây là phương trình đối xứng loại 1 , có trên mang nha .
câu d và e là phương trình đối xứng loại 2 , cũng có trên mạng nha .
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)