`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

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

\(B=\left(\frac{x}{2}+y\right)^3-6\left(\frac{x}{2}+y\right)^2.z+6\left(x+2y\right)z^2-8z^3\)

\(=\left(\frac{x}{2}+y\right)^3-3.\left(\frac{x}{2}+y\right)^2.2z+3.\left(\frac{x}{2}+y\right).\left(2z\right)^2-\left(2z\right)^3\)

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

\(C=\left(m-n\right)^3+15\left(m-n\right)^2.\left(m-p\right)-75\left(n-m\right)\left(p-m\right)^2-125\left(p-m\right)^3\)

\(=\left(m-n\right)^3+3.\left(m-n\right).\left[5\left(m-p\right)\right]+3.\left(m-n\right).\left[5\left(m-p\right)\right]^2+\left[5\left(m-p\right)\right]^3\)

\(=\left(m-n+5m-5p\right)^3=\left(6m-n-5p\right)^3\)

a, (x + y + z)(x - y - z)

= x^2 - xy - xz + xy - y^2 - zy + zx - zy - z^2

= x^2 + y^2 + z^2 + (xy - xy) + (xz - xz) - (zy + zy)

= x^2 + y^2 + z^2 - 2zy

b, (x - y + z)(x + y + z)

= x^2 + xy + xz - xy - y^2 - zy + zx + zy + z^2

= x^2 + y^2 + z^2 + (xy - xy) + xz + xz + (zy - zy)

= x^2 + y^2 + z^2 + 2zx

a) Ta có:

(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2

=x^2 - y^2 - 2yz - z^2.

b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2

=x^2 +2xz- y^2 +z^2.

c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)

= -16 + x^2 - 6x + 9

= x^2 - 6x - 7.

\(a,\left(x+y+z\right)\left(x-y-z\right)\)

\(=x\left(x-y-z\right)+y\left(x-y-z\right)+z\left(x-y-z\right)\)

\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)

\(=x^2-y^2-2yz-z^2\)

\(b,\left(x-y+z\right)\left(x+y+z\right)\)

\(=x\left(x+y+z\right)-y\left(x+y+z\right)+z\left(x+y+z\right)\)

\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)

\(=x^2+2xz-y^2+z^2\)

\(c,-16+\left(x-3\right)^2\)

\(=-16+x^2-6x+9\)

\(=x^2-6x-7\)

(x+y+z)(x+z-y)(x+y-z)(y+z-x)

=[(x+y)^2-z^2]*[(x+z-y)(y+z-x)]

=[(x+y)^2-z^2][y^2-(x+z)^2]

=(x^2+2xy+y^2-z^2][y^2-x^2-2xz-z^2]

=x^2y^2-x^4-2x^3z-x^2z^2+2xy^3-2x^3y-4x^2yz-2xyz^2+y^4-y^2x^2-2xy^2z-z^2y^2-y^2z^2+x^2z^2+2xz^3+z^4