giai he phuong trinh 2x^2-xy=xy^2_2x+y
(x^2+2y^2)(1+1/xy)^2=3
giai he phuong trinh
x+2\x+1\y=4
1\x^2+1\xy+x\y=3
giai he phuong trinh sau :
x^3 - x^2 y^2 - y^3 + 1 = 0 va x^3 + xy - 2 = 0
giai cac he phuong trinh sau
15) \(\left\{{}\begin{matrix}3x+2y=7\\x^2+y^2-7x+xy=0\end{matrix}\right.\)
16)\(\left\{{}\begin{matrix}2x+3y=5\\x^2+xy+y^2-4x=-1\end{matrix}\right.\)
>< giúp với ạ
Giai he phuong trinh : \(\left\{{}\begin{matrix}x^2+y^2+xy=3\\x^2+xy=7x+5y-9\end{matrix}\right.\)
giai he phuong trinh \(\left\{{}\begin{matrix}x^2+y^2+xy=1\\x^3+y^3=x+3y\end{matrix}\right.\)
HPT\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-xy=1-2xy\\\left(x+y\right)\left(1-2xy\right)=x+3y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=1\\x^2+xy=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=1\\y=-\sqrt{2};\sqrt{2}\end{matrix}\right.\)
The vao roi tinh la xong
giai he phuong trinh\(\left\{{}\begin{matrix}x^2+y^2-xy=19\\x+y+xy=-7\end{matrix}\right.\)
1) Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}x+y+xy=5\\x^2+y^2+x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^2-2xy+x+y=8\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^2+a-2b=8\end{matrix}\right.\) \(\Rightarrow a^2+a-2\left(5-a\right)=8\)
\(\Leftrightarrow a^2+3a-18=0\Rightarrow\left[{}\begin{matrix}a=3\Rightarrow b=2\\a=-6\Rightarrow b=11\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
giai he phuong trinh
\(\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\end{cases}}\)
\(\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(\frac{1}{y}+y\right)=\frac{9}{2}\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=5\end{cases}}\)
dat an phu r giai
giai phuong trinh xy+xz=2(x+y+z); xy+yz=3(x+y+z); xz+yz=4(x+y+z)
TH1:x,y,z=0
TH2:x=2\(\frac{3}{10}\)
y=3\(\frac{5}{6}\)
z=11\(\frac{1}{2}\)
giải ra cơ kết quả mik cx có mà hình như KQ sai rồi
à đúng rồi mà cách giải là sao v chỉ mik vs