10 tháng 11 2024 lúc 20:16
Các câu hỏi tương tự
\(\left\{{}\begin{matrix}x+y+3xy=-3\\xy+1=0\end{matrix}\right.\)
___
\(\left\{{}\begin{matrix}x^2-y^2=16\\x+y=8\end{matrix}\right.\)
giải hệ:
\(\left\{{}\begin{matrix}x+2y=7\\x^2+y^2-2xy=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y=2\\x^2+y^2+164\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y+xy=-13\\x^2+y^2-x-y=32\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y=3\\x^3-y^3=7\end{matrix}\right.\)
27 tháng 11 2023 lúc 21:00
Giải hpt:a)left{{}begin{matrix}dfrac{2y-5x}{3}+5dfrac{y+27}{4}-2xdfrac{x+1}{3}+ydfrac{6y-5x}{7}end{matrix}right.b)left{{}begin{matrix}dfrac{1}{2}left(x+2right)left(y+3right)-dfrac{1}{2}xy50dfrac{1}{2}xy-dfrac{1}{2}left(x-2right)left(y-2right)32end{matrix}right.c)left{{}begin{matrix}left(x+20right)left(y-1right)xyleft(x-10right)left(y+1right)xyend{matrix}right.d)left{{}begin{matrix}dfrac{2}{x+2y}+dfrac{1}{y+2x}3dfrac{4}{x+2y}-dfrac{3}{y+2x}1end{matrix}right.e)left{{}begin{matrix}dfrac{3x}{x+1}-df...
Đọc tiếp
Giải hpt:
a)\(\left\{{}\begin{matrix}\dfrac{2y-5x}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{y+2x}=3\\\dfrac{4}{x+2y}-\dfrac{3}{y+2x}=1\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
cho x,y,z thỏa mãn \(\left\{{}\begin{matrix}x^2+y^2+z^2=2\\xy+yz+xz=1\end{matrix}\right.\)
chứng minh \(\dfrac{-4}{3}\le x,y,z\le\dfrac{4}{3}\)
31 tháng 10 2021 lúc 20:50
GHPT
a) \(\left\{{}\begin{matrix}4x^2+1=y^2-4x\\x^2+xy+y^2=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+\dfrac{x+3y}{x^2+y^2}=3\\y-\dfrac{y-3x}{x^2+y^2}=0\end{matrix}\right.\)
22 tháng 10 2021 lúc 10:31
Giải hệ bằng phương pháp phân tích nhân tử
a) \(\left\{{}\begin{matrix}x^2+2y=xy+4\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-2xy+x+y=0\\x^4-4x^2y+3x^2+y^2=0\end{matrix}\right.\)
2 tháng 6 2023 lúc 16:56
\(\left\{{}\begin{matrix}x^3+xy^2+x^2+3x=2y^3+2x^2y+6y\\2\sqrt{y-1}+6\sqrt{xy-5x+3}=x^2+12x-16\end{matrix}\right.\)
19 tháng 5 2023 lúc 23:03
\(\left\{{}\begin{matrix}x^3+xy^2+x^2+3x=2y^3+2x^2y+6y\\2\sqrt{y-1}+6\sqrt{xy-5x+3}=x^2+12x-16\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+my=2\\mx-2y=1\end{matrix}\right.\)
tìm m để HPT có nghiệm (x;y) duy nhất thỏa mãn x<0 và y>0