bang 11

đọc đề bài đi ạ

\(a.\)

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

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

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

\(b.\)

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

Câu này đề sai sao ấy em !

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

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

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

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

`B=(x/2+y)^3-6(x/2+y)^2z + 6(x+2y)z^2-8z^3`

`=(x/2+y)^3 - 3. (x/2+y)^2 . 2z + 3. (x/2+y) . (2z)^2 - (2z)^3`

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

Sửa đề: Δ\(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)

Ta có: \(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)

\(=\left(\dfrac{1}{2}x+y\right)^2-3\cdot\left(\dfrac{1}{2}x+y\right)^2\cdot2z+3\cdot\left(\dfrac{1}{2}x+y\right)\cdot\left(2z\right)^2-\left(2z\right)^3\)

\(=\left(\dfrac{1}{2}x+y-2z\right)^3\)

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

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

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

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

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

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

Bài làm :

Ta có :

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

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

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

\(=\left(x-9\right)^2-\left(t+2\right)^2\)

Học tốt

x² - 6x + 5 - t² - 4t

= ( x² - 6x + 9 ) - ( t² + 4t + 4 )

= ( x - 3 )² - ( t + 2 )²

= [ ( x - 3 ) - ( t + 2 ) ][ ( x - 3 ) + ( t + 2 ) ]

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

= ( x - t - 5 )( x + t - 1 )

Phần in nghiêng mình viết thêm < nếu bạn cần >

bạn Nguyễn Thị Thu phải là (x-3)^2 chứ

=( 2x)² - 12x +9 -9 - y² + 2x +8

= (2x - 3)² - (y -1)²

^{k ai làm,tui làm}