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

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

\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2-3x^2\)

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

\(=y^2\)

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

Bài giải:

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

Bài làm

a) 2(x + y)³ + 2(x - y)³ 

= 2[(x + y)³ + (x - y)³]

= 2[x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³]

= 2[(x³ + x³) + (3x²y - 3x²y) + (3xy² + 3xy²) + (y³ - y³)]

= 2[2x³ + 6xy²]

= 4x³ + 12xy²

b)uhm... Mình sửa đề chút, thay vì là -3(x + y)²(x - y) thì mình sẽ thành +3(x + y)²(x - y)

(x - y)³ - (x + y)³ + 3(x + y)²(x - y) - 3(x + y)(x - y)²

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

= -[(x + y) - (x - y)]³ 

= -[x + y - x + y ]³

= -[y]³ 

= -y

a) \(x\left(x^2-16\right)-\left(x^2+1\right)\left(x-1\right)\) =\(x^3-16x^2-x^3+x^2-x+1\)

                                                        = \(x^2-17x+1\)

b) \(\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\) = \(\left(y^4-81\right)-\left(y^4-16\right)\)

                                                       =\(-65\)

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

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

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

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

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

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

\(=-z^2\)

\(\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)2xy}\)

=\(\frac{x^2+5x+y^2+5y+2xy-3}{x^2+6x+y^2+6y+2xy}\)

triệt tiêu x²;y²;2xy ta được:

\(\frac{5x+5y-3}{6x+6y}=\frac{5\left(x+y\right)-3}{6\left(x+y\right)}\)

=\(\frac{5.2010-3}{6.2010}=\frac{3349}{4020}\)

`(x+y+1)^3 - (x+y-1)^3 - 6(x+y)^2`

`=(x+y+1-x-y+1)[(x+y+1)^2 + (x+y+1)(x+y-1) + (x+y-1)^2] - 6(x+y)^2`

`=2(x^2+y^2 + 2xy+2x+2y + 1 + x^2 + 2xy +y^2 - 1 + x^2 + y^2 + 1 +2xy - 2x - 2y) - 6(x^2 + 2xy + y^2)`

`=2(3x^2 + 3y^2 + 6xy +1) - 6x^2 - 12xy - 6y^2`

`=6x^2 + 6y^2 + 12xy + 2 - 6x^2 - 12xy - 6y^2`

`=2`

\(=x^3+y^3+1-x^3-y^3+1-6\left(x^2+2xy+y^2\right)\\ =-6x^2-12xy+-6y^2+2\)

a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

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

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

Ta có: \(x^3-y^3+z^3+3xyz\)

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

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

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

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

TA có: \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\)

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

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

Ta có: \(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

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