Lời giải:

a. $x^2+y^2+4y+13-6x$

$=(x^2-6x+9)+(y^2+4y+4)$

$=(x-3)^2+(y+2)^2$

b.

$4x^2-4xy+1+2y^2-2y$

$=(4x^2-4xy+y^2)+(y^2-2y+1)$

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

c.

$x^2-2xy+2y^2+2y+1$

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

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

a. \(x^2+y^2+4y+12-6x=\left(x^2-6x+9\right)+\left(y^2+4y+4\right)=\left(x-3\right)^2+\left(y+2\right)^2\)b. \(4x^2-4xy+1+2y^2-2y=\left(4x^2-4xy+y^2\right)+\left(y^2-2y+1\right)=\left(2x-y\right)^2+\left(y-1\right)^2\)c. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)

a: \(x^2-6x+y^2+4y+13\)

\(=x^2-6x+9+y^2+4y+4\)

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

b: \(4x^2-4xy+1+2y^2-2y\)

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

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

c: \(x^2-2xy+2y^2+2y+1\)

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

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

a) x²+10x+26+y²+2y

=x²+10x+25+y²+2y+1

=(x+5)²+(y+1)²

b) z²-6z+5-t²-4t

=z²-6z+9-t²-4t-4

=(z-3)²-(t²+4t+4)

=(z-3)²-(t+2)²

c)x²-2xy+2y²+2y+1

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

d) 4x²-12x-y²+2y+8

=4x²-12x+9-y²+2y-1

=(2x-3)²-(y²-2y+1)

=(2x-3)²-(y-1)²

bạn ơi , bạn lấy bài này ở đâu vậy bạn

a)\(x^2+10x+26+y^2+2y\)

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

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

b)\(x^2-2xy+2y^2+1\)

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

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

c có lẽ sai ?

b:=y^2+2y+1+9x^2-12x+4

=(y+1)^2+(3x-2)^2

a:

SỬa đề: 5y^2

=y^2-10y+25+9x^2+4y^2-12xy

=(y-5)^2+(3x-2y)^2

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)

1) x² + 10x + 26 + y² + 2y 

= (x² + 10x + 25) + (y² + 2y + 1)

= (x² + 5x + 5x + 25) + (y² + y + y + 1)

= x(x + 5) + 5(x + 5) + y(y +  1) + (y + 1)

= (x + 5)² + (y + 1)²

2) z² - 6z + 13 + t² + 4t 

= (z² - 6z + 9) + (t² + 4t + 4) 

= (z² - 3z - 3z + 9) + (t² + 2t + 2t + 4)

= z(z - 3) - 3(z - 3) + t(t + 2) + 2(t + 2)

= (z - 3)² + (t + 2)²

3) x² - 2xy + 2y² + 2y + 1

(x² - 2xy + y²) + (y² + 2y + 1)

= (x - xy - xy + y²) + (y² + y + y +1)

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

= (x - y)² + (y + 1)²

a)  \(x^2-4x+5+y^2+2y=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)

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

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

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

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

bạn ơi , bạn lấy bài này ở đâu vậy bạn

tớ ko biết

1)a)x²+10x+26+y²+2y

=(x²+10x+25)+(y²+2y+1)

=(x+5)²+(y+1)²

b)x²-2xy+2y²+2y+1

=(x²-2xy+y²)+(y²+2y+1)

=(x-y)²+(y+1)²

c)z²-6z+13+t²+4t

=(z²-6z+9)+(t²+4t+4)

=(z-3)²+(t+2)²

d)4x²+2z²-4xz-2z+1

=(4x²-4xz+z²)+(z²-2z+1)

=(2x-z)²+(z-1)²

2)a)(x-3)²-4=0

<=>(x-3-2)(x-3+2)=0

<=>(x-5)(x-1)=0

<=>x-5=0 hoặc x-1=0

<=>x=5 hoặc x=1

b)x²-2x=24

<=>x²-2x-24=0

<=>(x²-6x)+(4x-24)=0

<=>x(x-6)+4(x-6)=0

<=>(x-6)(x+4)=0

<=>x-6=0 hoặc x+4=0

<=>x=6 hoặc x=-4

a) x^2 + 10x + 26 + y^2 + 2y

=x²+10x+25+y²+2y+1

=x²+2.x.5+5²+y²+2.y.1+1²

=(x+5)²+(y+1)²

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

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

c)z^2 - 6z + 13 + t^2 + 4t

=z²-6z+9+t²+4z+4

=z²-2.z.3+3²+t²+2.t.2+2²

=(z-3)²+(t+2)²

d)4x^2 + 2z^2 - 4xz - 2z + 1

=4x²-4xz+z²+z²-2z+1

=(2x)²-2.2x.z+z²+z²-2z.1+1²

=(2x-z)²+(z-1)²

a) x^2 + 10x + 26 + y^2 + 2y

=x²+10x+25+y²+2y+1

=x²+2.x.5+5²+y²+2.y.1+1²

=(x+5)²+(y+1)²

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

=x²-2xy+y²+y²+2y+1

=(x-y)²+(y+1)²

c)z^2 - 6z + 13 + t^2 + 4t

=z²-6z+9+t²+4z+4

=z²-2.z.3+3²+t²+2.t.2+2²

=(z-3)²+(t+2)²

d)4x^2 + 2z^2 - 4xz - 2z + 1

=4x²-4xz+z²+z²-2z+1

=(2x)²-2.2x.z+z²+z²-2z.1+1²

=(2x-z)²+(z-1)²

a) (x-3)^2 - 4 =0

<=>(x-3)²-2²=0

<=>(x-3-2)(x-3+2)=0

<=>(x-5)(x-1)=0

<=>x-5=0 hoặc x-1=0

<=>x=5 hoặc x=1

b) x^2 - 2x = 24

<=>x²-2x-24=0

<=>x²-2x+1-25=0

<=>(x-1)²-25=0

<=>(x-1)²-5²=0

<=>(x-1-5)(x-1+5)=0

<=>(x-6)(x+4)=0

<=>x-6=0 hoặc x+4=0

<=>x=6 hoặc x=-4

\(x^2+10x+26+y^2+2y=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)=\left(x+5\right)^2+\left(y+1\right)^2\)