\(a.\)

\(z^2-6z+5-t^2-4t\)

\(=z^2-6z+9-\left(t^2+4t+4\right)\)

\(=\left(z-3\right)^2-\left(t+2\right)^2\)

\(b.\)

\(4x^2-12x-y^2+2y+1\)

Câu này đề sai sao ấy em !

b, mik nghĩ đề sửa thành: \(4x^2-12x-y^2+2y+8\)

\(=4x^2-12x+9-y^2+2y-1\)

\(=\left(2x\right)^2-2.2.3.x+3^2-\left(y^2-2y+1\right)\)

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

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

`=x^2-2x+1+4y^2+4y+1`

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

`b)4x^2+y^2+12x+4y+13`

`=4x^2+12x+9+y^2+4y+4`

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

`c)x^2+17+4y^2+8x+4y`

`=x^2+8x+16+4y^2+4y+1`

`=(x+4)^2+(2y+1)^2`

`d)4x^2-12xy+y^2-4y+13`

`=4x^2-12x+9+y^2-4y+4`

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

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

b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)

c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)

d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)

a: \(x^2-2x+2+4y^2+4y\)

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

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

b: \(4x^2+12x+y^2+4y+13\)

\(=4x^2+12x+9+y^2+4y+4\)

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

c: \(x^2+8x+4y^2+4y+17\)

\(=x^2+8x+16+4y^2+4y+1\)

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

d: \(4x^2-12x+y^2-4y+13\)

\(=4x^2-12x+9+y^2-4y+4\)

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

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

b, \(\left(y+2z-3\right)\left(y-2z-3\right)=\left(y-3+2z\right)\left(y-3-2z\right)=\left(y-3\right)^2-\left(2z\right)^2\)

c, \(\left(x-y-6\right)\left(x+y-6\right)=\left(x-6-y\right)\left(x-6+y\right)=\left(x-6\right)^2-y^2\)

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

\(x^2-2xy+5y^2+4y+1\)

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

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

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

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

\(a,x^3+6x^2y+12xy^2+8y^3\\ =x^3+3.2x^2+3.2^2.x+\left(2y\right)^3\\ =\left(x+2y\right)^3\)

\(b,x^3-3x^2+3x-1\\ =x^3-3x^2.1+3x.1^2-1^3\\ =\left(x-1\right)^3\)

a) \(x^3+6x^2y+12xy^2+8y^3\)

\(=x^3+3\cdot x^2\cdot2y+2\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)

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

b) \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

\(=\left(x-1\right)^3\)

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) 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

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: \(x^2-2xy+2y^2+2y+1\)

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

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

\(1.z^2-6z+5-t^2-4t\)

\(=\left(z^2-6z+9\right)-\left(t^2+4t+4\right)\)

\(=\left(z-3\right)^2-\left(t+2\right)^2\)

\(3,x^2-2xy+2y^2+2y+1\)

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

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