\(x^2+2y^2-2xy+2x-4y+2=0\)
\(\Rightarrow x^2-2xy+y^2+2\left(x-y\right)+1+y^2-2y+1=0\)
\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-1\right)^2=0\)
\(\Rightarrow\left(x-y+1\right)^2+\left(y-1\right)^2=0\)
=>................
Giải chi tiêt hộ mk.
Chứng minh rằng với mọi x,y ta luôn có:
√((x^2+4y^2)/2)+√((x^2+2xy+4y^2)/3)\(\ge\)x+2y.
giải hệ :1, x^3-6x^2y+9xy^2-4y^3=0 v căn (x-y) + căn (x+y) = 2
2,xy+x-2=0 v 2x^3-x^2y+x^2+y^2-2xy-y=0
1) 8y^2-25=3xy+5x
2)xy-2y-3=3x-x^2
3)x^2+2y^2-3xy_4x-3y-26=0
4)x^2+3y^2+2xy-2x-4y-3=0
5)x^3+3x=y^3
6)x^4-2x^2y+7y^2=55
7)x^2y^2-2xy=x^2+16y^2
Mọi người giúp toy với , thanks with love :>
\(\hept{\begin{cases}x^4+x^2y^2+2x^2-2x^2y-2y^3-4y=0\\2x^2+y^2+x=14\end{cases}}\)
Giải hệ \(\left\{{}\begin{matrix}xy+3y^2-x+4y=7\\2xy+y^2-2x-2y+1=0\end{matrix}\right.\)
\(P=x^4+y^4+x^4y^4+1=\left(\left(x+y\right)^2-2xy\right)^2-2x^2y^2+x^4y^4+1\)
\(=\left(10-2xy\right)^2-2x^2y^2+x^4y^4+1=x^4y^4+2x^2y^2-40xy+101\)
\(=\left(x^2y^2-4\right)^2+10\left(xy-2\right)^2+45\ge45\)
Dấu bằng tự xét
1. Cho x,y,z >0 t/m: \(\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}=2\)
Tìm max (xyz)
2. Cho \(2x^2+y^2-2xy=1\)
a) CM: |x| ≤ 1
b) Tìm max \(P=4x^4+4y^4-2x^2y^2\)
\(\hept{\begin{cases}x^4+6x^2y+3xy^2+2xy+y^4+4y^2=x^3+6x^2y^2+4x^2+x+2y^2+4y\\4x^3y+6xy^2+4x+y^3+y^2+13=2x^3+3x^2y+x^2+4xy^3+8xy+y\end{cases}}\)
\(\hept{\begin{cases}3x^2+2y+1=2z\left(x+2\right)\\3y^2+2z+1=2x\left(y+2\right)\\3z^2+2x+1=2y\left(z+2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}3x^2+2y+1=2xz+4z\\3y^2+2z+1=2xy+4x\\3z^2+2x+1=2yz+4y\end{cases}}}\)
Cộng 3 vế vào rồi chuyển vế ta được
\(2x^2+2y^2+2z^2-2xy-2yz-2zx+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2 +\left(z-x\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
Dễ thấy VP > 0
Dấu "=" khi x = y = z = -1
