\(a.\)

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

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

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

\(b.\)

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

Câu này đề sai sao ấy em !

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

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

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

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

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) 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^2-2x+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) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)

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

b) \(x^2+y^2+2xy=\left(x+y\right)^2\)

c) \(9x^2+12x+4=\left(3x+2\right)^2\)

d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)

Bài 1:

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) \(z^2-6z+5-t^2-4t=(z^2-6t+9)-(t^2+4t+4)\)

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

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

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

d) \(4x^2-12x-y^2+2y+8=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)

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

Bài 2:

a) \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-16\)

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

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

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

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

1a/ z² - 6z + 5 - t² - 4t = z² - 2 . 3z + 3² - 4 - t² - 4t = (z² - 2 . 3z + 3²) - (2² + 2 . 2t + t²) = (z - 3)² - (2 + t)²

b/ x² - 2xy + 2y² + 2y² + 1 = x² - 2xy + y² + y² + 2y + 1 = (x² - 2xy + y²) + (y² + 2y + 1) = (x - y)² + (y + 1)²

c/ 4x² - 12x - y² + 2y + 8 = (2x)² - 12x - y² + 2y + 3² - 1 = [ (2x)² - 2 . 3 . 2x + 3² ] - (y² - 2y + 1) = (2x - 3)² - (y - 1)²

2a/ (x + y + 4)(x + y - 4) = x² + xy - 4x + xy + y² - 4y + 4x + 4y + 16 = x² + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y² + 16

= x² + 2xy + y² + 4² = (x + y)² + 4²

b/ (x - y + 6)(x + y - 6) = x² + xy - 6x - xy - y² + 6y + 6x + 6y - 36 = x² + (xy - xy) + (-6x + 6x) + (6y + 6y) - y² - 36

= x² - y² + 12y - 6² = x² - (y² - 12y + 6²) = x² - (y² - 2 . 6y + 6²) = x² - (y - 6)²

c/ (y + 2z - 3)(y - 2z - 3) = y² -2yz - 3y + 2yz - 4z² - 6z - 3y + 6z + 9 = y² + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z² + 9

= y² - 6y - 4z² + 9 = (y² - 6y + 9) - 4z² = (y - 3)² - (2z)²

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x² + 4y² + 6yz - 2xy + 6yz + 9z² - 3xz = (2xy - 2xy) + (3xz - 3xz) - x² + (6yz + 6yz) + 9z² + 4y²

= -x² + 4y² + 12yz + 9z² = (4y² + 12yz + 9z²) - x² = [ (2y)² + 2 . 2 . 3yz + (3z)² ] - x² = (2y + 3z)² - x²

:v dễ mà có trong nâng cao mới hc qua :3

a, x2+10x+26+y2+2y

=(x2+2.x.5+52)+(12+2.1.y+y2)

=(x+5)2+(y+1)2

b, x2−2xy+2y2+2y+1

=x2−2xy+y2+y2+2y+1

=(x2−2.x.y+y2)+(y2+2.y.1+12)

=(x−y)2+(y+1)2

c,z2−6z+5−t2−4t

=−(t2+4t−z2+6z−5)

=−(t2+2.t.2+22−z2+2.z.3−32)

=−((t2+2.t.2+22)−(z2−2.z.3+32))

=−((t+2)2−(z−3)2)

=(z−3)2−(t+2)2

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

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

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

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

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

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