Tìm x, y nguyên:
x^2 -2x -11 =y^2
x^2 + 2xy + 7(x+y) + 2y^2 +10=0
Tìm số nguyên x biết
a,3x+3y-2xy=7
b,xy+2x+y+11=0
c,xy+x-y=4
d,2x.(3y-2)+(3y-2)=12
e,3x+4y-xy=15
f,xy+3x-2y=11
g,xy+12=x+y
h,xy-2x-y=-6
i,xy+4x=25+5y
ii,2xy-6y+x=9
iii,xy-x+2y=3
k,2.x^2.y-x^2-2y-2=0
l,x^2.y-x+xy=6
Tìm x,y biết:
a,2x^2+y^2+2xy+10x+25=0
b,x^2+3y^2+2xy-2y+1=0
c,x^2+2y^2+2xy-2x+2=0
a) \(2x^2+y^2+2xy+10x+25=0\)
\(\Leftrightarrow x^2+x^2+y^2+2xy+10x+25=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+10x+25\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+5\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x\\\left(x+5\right)^2\ge0\forall x\end{cases}}\)
\(\Rightarrow\left(x+y\right)^2+\left(x+5\right)^2\ge0\forall x\)
Vậy đẳng thức xảy ra\(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=5\end{cases}}\)
b)\(x^2+3y^2+2xy-2y+1=0\)
\(\Leftrightarrow x^2+y^2+2y^2+2xy-2y+\frac{1}{2}+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(2y^2-2y+\frac{1}{2}\right)+\frac{1}{2}=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)
Vì \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2\ge0\)
nên \(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)
Mà\(\left(x+y\right)^2+\left(\sqrt{2}y-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}=0\)
nên pt vô nghiệm
a) 2x2 + y2 + 2xy + 10x + 25 = 0
=> (x2 + 2xy + y2) + (x2 + 10x + 25) = 0
=> (x + y)2 + (x + 5)2 = 0
<=> \(\hept{\begin{cases}x+y=0\\x+5=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x\\x=-5\end{cases}}\) <=> \(\hept{\begin{cases}y=5\\x=-5\end{cases}}\)
b)c) xem lại đề
(x-5).y+2x-10=7
2.(3-y)-x(y-3)=19
xy-x=2y+9
xy-3y+6x=17+2x2
x2y+4x=11+2xy
GPT: 9x/(2x2 + x + 3) - x/(2x2 - x - 3) = 8
Tìm các số nguyên x,y thỏa mãn: x2 + 2xy + 7(x+y) + 2y2 + 10= 0
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
1)\(\begin{cases}x^4+2xy+6y-7x^2-2x^2y+9=0\\2x^2y-x^3=10\end{cases}\)
2)\(\begin{cases}2x^3-x^2y+x^2+y^2-2xy-y=0\\xy+x-2=0\end{cases}\)
\(\text{tìm x,y 2x^2 + y^2 - 2xy - 4x - 2y + 10 = 0}\)
tìm các cặp số (x,y) t/m \(2x^2+2y^2-2xy+y+x-10=0\)
tìm x,y thuộc z để
2x^2+2xy+y^2-4x+2y+10=0