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`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: \(B=\left(\dfrac{x}{2}+y\right)^3-6\left(\dfrac{x}{2}+y\right)^2z+12\left(x+2y\right)\cdot z^2-8z^3\)
\(=\left(\dfrac{1}{2}x+y\right)^2-3\cdot\left(\dfrac{1}{2}x+y\right)^2\cdot2z+3\cdot\left(\dfrac{1}{2}x+y\right)\cdot\left(2z\right)^2-\left(2z\right)^3\)
\(=\left(\dfrac{1}{2}x+y-2z\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z+x\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a . \(\left(x+y+4\right)\left(x+y-4\right)=\left(x+y\right)^2-4^2\)
b . \(\left(x-y+6\right)\left(x+y-6\right)=x^2-\left(y-6\right)^2\)
c . \(\left(y+2z-3\right)\left(y-2z-3\right)=\left(y-3\right)^2-\left(2z\right)^2\)
d . \(\left(x+2y+3z\right)\left(2y+3z-x\right)=\left(2y+3z\right)^2-x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a///(y+2z-3)(y-2z-3)
= y+2z-3 * y-2z-3
= y2- 32
b///(x+2y+3z)(3y+3z-x)
=x+2y+3z * 3y+3z-x
=6y2+6z
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2+10xx+26+y^{22}+2\)
Nhóm và rút nhân tử chung là UCLN ra ngoài, sau đó kết hợp:
\(11x^2+26+y^{22}+2y\)
c) \(4x^2-12x^2-y^2+2y^2+1\)
=\(-8x^2+y^2+1\)
d) \(\left(y+2z-3\right)\left(y-3-2z\right)=\left[\left(y-3\right)+2z\right]\left[\left(y-3\right)-2z\right]\)
\(=\left(y-3\right)^2-4z^2\)
x2 + 10x + 26 + y2 + 2y
= ( x2 + 10x + 25 ) + ( y2 + 2y + 1 )
= ( x + 5 )2 + ( y + 1 )2
4x2 - 12x - y2 + 2y + 1 ( thiếu đề hay sao ý )
( y + 2z - 3 )( y - 3 - 2z )
= [ y - ( 3 - 2z ) ][ ( y + ( 3 - 2z ) ]
= y2 - ( 3 - 2z )2
a)
Ta có :
\(\left(y+2z-3\right)\left(y-2z-3\right)\)
\(=\left[\left(y-3\right)+2z\right]\left[\left(y-3\right)-2z\right]\)
\(=\left(y-3\right)^2-2z^2\)
b)
Ta có :
\(\left(x-y+6\right)\left(x+y-6\right)\)
\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)
\(=x^2-\left(y-6\right)^2\)
thíu ngoặc