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

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

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

a: \(x^2+3y^2-4x+6y+7=0\)

\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)

a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)

\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)

b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)

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

c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)

\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)

`x^2 -4x+4-y^2`

`=(x^2 -4x+4)-y^2`

`=(x-2)^2 -y^2`

`=(x-2-y)(x-2+y)`

`x^2+2xy+y^2-x-y`

`=(x^2+2xy+y^2) -(x+y)`

`=(x+y)^2 -(x+y)`

`=(x+y)(x+y-1)`

`x^2-2xy+y^2-9`

`=(x^2-2xy+y^2)-3^2`

`=(x-y)^2-3^3`

`=(x-y-3)(x-y+3)`

Tách ra đi cậu.

https://hoc24.vn/cau-hoi/927x118.8505894378996

giúp mik với ạ

a)2x^2+xy-y^2-x+2y-1

=2x^2+xy-x-(y-1)^2

=2x^2+x(y-1)-(y-1)^2

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

=(2x-y+1)(x+y-1)

a: \(P=x^2+y^2-6x-2y+17\)

\(=x^2-6x+9+y^2-2y+1+7\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+7\ge7\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-1=0

=>x=3 và y=1

b: \(Q=x^2+xy+y^2-3x-3y+999\)

\(=x^2+x\left(y-3\right)+y^2-3y+999\)

\(=x^2+2\cdot x\cdot\left(\frac12y-\frac32\right)+\left(\frac12y-\frac32\right)^2+y^2-3y-\left(\frac12y-\frac32\right)^2+999\)

\(=\left(x+\frac12y-\frac32\right)^2+y^2-3y-\left(\frac14y^2-\frac32y+\frac94\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34y^2-\frac32y-\frac94+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y-3\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y^2-2y+1-4\right)+999\)

\(=\left(x+\frac12y-\frac32\right)^2+\frac34\left(y-1\right)^2+996\ge996\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x+\frac12y-\frac32=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=-\frac12y+\frac32=-\frac12+\frac32=\frac22=1\end{cases}\)

c: \(R=2x^2+2xy_{}+y^2-2x+2y+15\)

\(=x^2-4x+4+x^2+2xy+y^2+2x+2y+11\)

\(=\left(x-2\right)^2+x^2+2xy+y^2+2x+2y+1+10\)

\(=\left(x-2\right)^2+\left(x+y+1\right)^2+10\ge10\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y+1=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x-1=-2-1=-3\end{cases}\)

d: \(S=x^2+26y^2-10xy+14x-76y+59\)

\(=x^2-10xy+25y^2+14x-70y+y^2-6y+59\)

\(=\left(x-5y\right)^2+14\left(x-5y\right)+49+y^2-6y+9+1\)

\(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-3=0\\ x-5y+7=0\end{cases}\Rightarrow\begin{cases}y=3\\ x=5y-7=5\cdot3-7=15-7=8\end{cases}\)

e: \(T=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+4y^2+10x-20y+y^2-2y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y-1=0\\ x-2y+5=0\end{cases}\Rightarrow\begin{cases}y=1\\ x=2y-5=2\cdot1-5=2-5=-3\end{cases}\)