\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

\(3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{25}{8}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x+\dfrac{5}{2}\right)^2+\dfrac{183}{8}=0\\ \Leftrightarrow x,y\in\varnothing\)

Sửa đề: \(3x^2+6y^2-12x-20y+40=0\)

\(\Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+\dfrac{50}{3}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)

\(2\left(x^2+y^2\right)=\left(x+y\right)^2\\ \Leftrightarrow2x^2+2y^2=x^2+2xy+y^2\\ \Leftrightarrow x^2-2xy+y^2=0\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x-y=0\Leftrightarrow x=y\)

xy là x.y hay là x và y vậy bn

X và y là số nguyên phải ko

Thiếu đề !

Chuyển biểu thức , ta có :

x² - 8xy + 20y² - 4y + 1

= x² - 8xy + 16y² + 4y² - 4y + 1

= (x² - 8xy + 16y²) + (4y² - 4y + 1)

= (x - 4y)² + (2y + 1)² 

Còn lại do thiếu đề nên không thể làm tiếp 

Ah, mình đánh thiếu mất  đề là \(x^2-8xy+20y^2-4y+1=0\)

d) \(4x^2y^2-8xy^2+4y^2=4y^2\left(x^2-2x+1\right)=4y^2\left(x-1\right)^2\)

e) \(x^3y+10x^2y+35xy=xy\left(x^2+10x+35\right)\)

f) \(2x^3-4x^2y+2xy^2-8x=2x\left(x^2-2xy+y^2-4\right)=2x\left[\left(x-y\right)^2-4\right]=2x\left(x-y-2\right)\left(x-y+2\right)\)

g) \(3x^2-9xy-6x+18y=3x\left(x-3y\right)-6\left(x-3y\right)=3\left(x-3y\right)\left(x-2\right)\)

h) \(x^2y^2-3xy^2+2xy-6y=xy\left(xy+2\right)-3y\left(xy+2\right)=\left(xy+2\right)\left(xy-3y\right)=y\left(xy+2\right)\left(x-3\right)\)

d: \(4x^2y^2-8xy^2+4y^2\)

\(=4y^2\left(x^2-2x+1\right)\)

\(=4y^2\left(x-1\right)^2\)

e: \(x^3y+10x^2y+35xy\)

\(=xy\left(x^2+10x+35\right)\)

f: \(2x^3-4x^2y+2xy^2-8x\)

\(=2x\left(x^2-2xy+y^2-4\right)\)

\(=2x\left(x-y-2\right)\left(x-y+2\right)\)

g: \(3x^2-9xy-6x+18y\)

\(=3x\left(x-3y\right)-6\left(x-3y\right)\)

\(=3\left(x-2\right)\left(x-3y\right)\)

h: \(x^2y^2-3xy^2+2xy-6y\)

\(=xy^2\left(x-3\right)+2y\left(x-3\right)\)

\(=y\left(xy+2\right)\left(x-3\right)\)

pặc pặc....pặc pặc...........pặc pặc......

._.

\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

Với \(x=0\Leftrightarrow y=0\), 

Với \(x,y\ne0\): 

\(\left(\sqrt{x^2+1}-x\right)\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\sqrt{x^2+1}-x\)

\(\Leftrightarrow y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\)

Tương tự ta cũng có: \(x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\)

suy ra \(x+y=-\left(x+y\right)\Leftrightarrow x+y=0\)

\(M=10x^4+8y^4-15xy+6x^2+5y^2+2017\)

\(=18x^4+26x^2+2017\ge2017\)

Dấu \(=\)tại \(x=0\Rightarrow y=0\).

a: Sửa đề: \(2A+\left(2x^2+y^2\right)=6x^2+5y^2-2x^2y^2\)

=>\(2A=6x^2+5y^2-2x^2y^2-2x^2-y^2\)

=>\(2A=4x^2+4y^2-2x^2y^2\)

=>\(A=2x^2+2y^2-x^2y^2\)

b: \(2A-\left(xy+3x^2-2y^2\right)=x^2-8y+xy\)

=>\(2A=x^2-8y+xy+xy+3x^2-2y^2\)

=>\(2A=4x^2+2xy-8y-2y^2\)

=>\(A=2x^2+xy-4y-y^2\)

c: Sửa đề: \(A+\left(3x^2y-2xy^2\right)=2x^2y+4xy^3\)

=>\(A=2x^2y+4xy^3-3x^2y+2xy^2\)

=>\(A=-x^2y+4xy^3+2xy^2\)