Đề bài là gì sao không ghi rõ?? 

a,  x²+2xy+y²+2x+2y-15

<=> (x+y )²+2(x+y)+1-16

Đặt x+y =a

<=> a²+2a+1-4²

<=> (a+1)²-4²

<=> (a+5)(a-3) =>( x+y+5)(x+y-3)

b, x²-4xy+4y²-2x-4y-35

<=> (x-2y)²-2(x-2y)+1-36

Đặt (x-2y)  =b 

=> b²-2b+1-6²

<=> (b-1)²-6²

<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)

c, 

a,A= x^2+2xy+y^2+2x+2y-15

= (x+y)^2+(x+y)-15

Đặt x+y=a, ta có:

A=a^2+2a-15

  =a^2+2a+1-16

  =(a+1)^2-4^2

  =(a+1+4)(a+1-4)

  =(a+5)(a-3)

Thay a=x+y, ta có: A=(x+y+5)(x+y-3).

\(B=3x^2-5x+7=3\left(x-\frac{5}{6}\right)^2+\frac{59}{12}\ge\frac{59}{12}\)

\(C=x^2-4x+3+11=\left(x^2-4x+4\right)+10=\left(x-2\right)^2+10\ge10\)

\(D=-x^2-4x-y^2+2y=-\left(x^2-4x+4\right)-\left(y^2-2y+1\right)+5=-\left[\left(x-2\right)^2+\left(y-1\right)^2\right]+5\le5\)

a=(2x+y)^2+(x-1)^2+(y+2)^2+2021-5=2016

Amin=2016

Ta có: \(5x^2-4xy+2x-2y+y^2+2=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4x-2y\right)+1+\left(x^2-2x+1\right)==0\)

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

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

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(2x-y+1\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

y sai rùi bn

\(x^2+4y^2-5x+10y-4xy+20\)

\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)

\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)

\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được : 

\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)

\(B=x^2-2xy-2x+2y+y^2\)

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

\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được : 

\(=1-2=-1\)

a) \(\Leftrightarrow x^2-2xy+y^2+y^2-4y+4+1\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)

 vậy Min=1 khi x-y =0 , y-2 = 0 <=> x=y,y=2=>x=y=2

|Mấy câu sau tương tự nếu ko biết thì nói nha

,

a: =5x^2(y+1)-4z(y+1)

=(y+1)(5x^2-4z)

b: =2x(2y-1)-z(2y-1)

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

a) \(5x^2y-4zy+5x^2-4z\)

\(=\left(5x^2y+5x^2\right)-\left(4zy+4z\right)\)

\(=5x\left(y+1\right)-4z\left(y+1\right)\)

\(=\left(y+1\right)\left(5x-4z\right)\)

b) \(4xy-2x-2yz+z\)

\(=\left(4xy-2x\right)-\left(2yz-z\right)\)

\(=2x\left(2y-1\right)-z\left(2y-1\right)\)

\(=\left(2y-1\right)\left(2x-z\right)\)