tim x , y :
/ x + 3y -1 / + / 2y - \(\frac{1}{2}\) /^200 = 0
Những câu hỏi liên quan
5/ Tim x,y,z biet
a/x^2+2y^2+2xy-2y+1=0
b/5x^2+3y^2+2^2-4x+6xy+4z+6=0
a)\(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)
Vậy x=-1 y=1
Đúng 0
Bình luận (0)
a) \(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)
b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)
\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow\) \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
\(\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow y=-x=-1\)
\(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)
Đúng 0
Bình luận (0)
\(\frac{1+y}{9}=\frac{1+2y}{7}=\frac{1+3y}{x}\) tim x ?
tim nghiem nguyen cua phuong trinh
x^3-y^3-2y^2-3y-1=0
\(x^3-y^3-2y^2-3y-1=0\)
\(<=>x^3=y^3+2y^2+3y+1\)≤\(y^3+3y^2+3y+1=(y+1)^3\)(vì \(y^2\)≥0) (1)
Ta có:\(x^3=y^3+2y^2+3y+1>y^3-3y^2+3y-1\)\(=(y-1)^3\) (2)
Từ (1) và (2)
\(=>(y-1)^3< y^3+2y^2+3y+1=x^3 =<(y+1)^3\)
\(=>y^3+2y^2+3y+1=y^3,(y+1)^3\)
Xong giải ra thôi
Đúng 0
Bình luận (0)
Rất xin lỗi bạn vì đến năm 2021 bn ms nhận được câu trả lời
Đúng 0
Bình luận (0)
Giải các hệ phương trình:
\(a,\left\{{}\begin{matrix}\frac{3x-2y}{5}+\frac{5x-3y}{3}=x+1\\\frac{2x-3y}{3}+\frac{4x-3y}{2}=y+1\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}\frac{1}{x-3}-\frac{1}{y-1}=0\\3x-2y=7\end{matrix}\right.\)
1)begin{cases}x^2-yleft(x+yright)+10left(x^2+1right)left(x+y-2right)+y0end{cases}2)begin{cases}x^2-4x+y^4+4y^22xy^2+2y^2+6x23end{cases}3)begin{cases}2x+frac{1}{x+y}34x^2+4y^2+4xy+frac{3}{left(x+yright)^2}7end{cases}4)begin{cases}y^6+x^9+3y^4+3y^284y^2-3x^3y^2+x^32end{cases}5)begin{cases}sqrt{x+y}-2sqrt{x-y}1x+sqrt{x^2+y^2}8end{cases}6) begin{cases}x+y-2frac{y}{x^2+1}x^2+y^2+xyy-1end{cases}7) begin{cases}4x-1sqrt{left(2x+yright).left(2y+1right)}sqrt{x+2y+1}-sqrt{x+y-1}sqrt{x-1}end{cases}8) begin{...
Đọc tiếp
1)\(\begin{cases}x^2-y\left(x+y\right)+1=0\\\left(x^2+1\right)\left(x+y-2\right)+y=0\end{cases}\)
2)\(\begin{cases}x^2-4x+y^4+4y^2=2\\xy^2+2y^2+6x=23\end{cases}\)
3)\(\begin{cases}2x+\frac{1}{x+y}=3\\4x^2+4y^2+4xy+\frac{3}{\left(x+y\right)^2}=7\end{cases}\)
4)\(\begin{cases}y^6+x^9+3y^4+3y^2=8\\4y^2-3x^3y^2+x^3=2\end{cases}\)
5)\(\begin{cases}\sqrt{x+y}-2\sqrt{x-y}=1\\x+\sqrt{x^2+y^2}=8\end{cases}\)
6) \(\begin{cases}x+y-2=\frac{y}{x^2+1}\\x^2+y^2+xy=y-1\end{cases}\)
7) \(\begin{cases}4x-1=\sqrt{\left(2x+y\right).\left(2y+1\right)}\\\sqrt{x+2y+1}-\sqrt{x+y-1}=\sqrt{x-1}\end{cases}\)
8) \(\begin{cases}\left(x+y\right).\left(x+4y^2+y\right)+3y^4=0\\\sqrt{x+2y^2+1}-y^2+y+1=0\end{cases}\)
ôi trờiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Đúng 0
Bình luận (0)
Cho bieu thuc
\(P=\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(x+1\right)}-\frac{x^2y^2}{\left(x+1\right)\left(1-y\right)}\)
Rut gon P= x+xy-y
DKXD \(x\ne y\); \(x\ne-1\):\(y\ne1\)
Tim x y de P nguyen duong thoa man \(x^2+y^2+3xy-x-3y=0\)
cho x,y>0 và xy=1. Tim GTLN A=x^2+3x+y^2+3y+\(\frac{9}{x^2+y^2+1}\)
Tìm đa thức M , biết :
a) \(M-\left(\frac{1}{2}x^2y-5xy^2+x^3-y^3\right)=\frac{3}{4}xy^2-2x^2y+\)\(2y^3-\frac{1}{3}x^3\)
b)\(\left(-\frac{1}{3}x^3y^3+5x^2y^2-\frac{5}{2}xy\right)-M=xy-\frac{1}{6}x^3y^3-3x^2y^2\)
c)\(\left(\frac{2}{7}xy^4-5x^5+7x^2y^3-3\right)+M=0\)
tim x,y,z trong cac truog hop sau
a)2x=3y=5z va |x+2y|=5
b)5x=2y;2x=3z va xy=90
c) \(\frac{y+z+1}{x}\)=\(\frac{x+z+2}{y}\)=\(\frac{x+y-3}{z}\)=\(\frac{1}{x+y+z}\)
a)
\(2x=3y\Rightarrow y=\frac{2x}{3}\)
\(!x+2y!=5\Rightarrow\orbr{\begin{cases}x+2y=5\\x+2y=-5\end{cases}\Rightarrow\orbr{\begin{cases}x+2.\frac{2}{3}x=5\Rightarrow x=\frac{15}{7}\\x+2.\frac{2}{3}x=-5\Rightarrow x=-\frac{15}{7}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}y=\frac{10}{7}\\y=\frac{-10}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}z=\frac{6}{7}\\z=\frac{6}{7}\end{cases}}\)
(x,y,z)=(15/7,10/7,6/7)
(x,y,z)=(-15/7,-10/7,-6/7)
Đúng 0
Bình luận (0)