a: =(x+y)^3+z^3-3xy(x+y)-3xyz

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

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

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

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

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

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

c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x-1)

d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2

=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c

=b^2(c-a)+b(c^2-a^2)+ac(c-a)

=(c-a)(b^2+ac)+b(c-a)(c+a)

=(c-a)(b^2+ac+bc+ba)

=(c-a)[b^2+bc+ac+ab]

=(c-a)[b(b+c)+a(b+c)]

=(c-a)(b+c)(b+a)

( x + y + z)³ - x³ - y³ - z³=x³+y³+z³+3(a+b)(a+c)(b+c)- x³ - y³ - z³

                                              = 3(a+b)(b+c)(a+c)

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

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

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

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

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

\(=3\left(x+y\right)\left[xy-z\left(x+y-z\right)\right]\)

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

\(=3\left(x+y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)

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

#\(Urushi\text{☕}\)

Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:

(x+y+z)3-x3-y3-z3

=[(x+y)+z]3-x3-y3-z3

=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3

=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3

=3(x+y)(xy+xz+yz+z2)

=3(x+y)[x(y+z)+z(y+z)]

=3(x+y)(y+z)(x+z)

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

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

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

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

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

=3(y-z)(x+y)(x-z)

Ta có: ( x - y) z³ + ( y - z ) x³ + ( z - x ) y³  

= ( x - y ) z³ + ( y - z )x³ + ( z - y)y³ + ( y - x ) y³

= ( x - y ) ( z³ - y³ ) + ( y - z ) ( x³ - y³) 

= ( x - y ) ( z - y ) ( z² + zy + y² ) + ( y - z ) ( x - y) ( x² + xy + y² ) 

= ( x - y ) ( y - z ) ( x² + xy + y² - z² - zy - y²) 

= ( x - y ) ( y - z ) [ ( x² - z²) + ( xy - zy) ]

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

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

(x - y).z³ + (y - z).x³ + (z - x).y³

= z³(x - y) + x³y - x³z + y³z - xy³

= z³(x - y) + xy(x² - y²) - z(x³ - y³)

= z³(x - y) + xy(x - y)(x + y) - z(x - y)(x² + xy + y²)

= (x - y)(z³ + x²y + xy² - x²z - xyz - y²z)

= (x - y)[z(z² - x²) + xy(x - z) + y²(x - z)]

= (x - y)[z(z - x)(z  + x) - xy(z- x) - y²(z - x)]

= (x - y)(z - x)(z² + xz - xy - y²)

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

= (x  - y)(z - x)(y - z)(y + z - x)

`a, x^3 + y^3 + x + y`

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

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

`b, x^3 - y^3 + x -y`

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

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

`c, (x-y)^3 + (x+y)^3`

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

`= (2x)(x^2 + 3y^2)`

`d, x^3 - 3x^2y + 3xy^2 - y^3 + y^2 - x^2`

`= (x-y)^3 + (y-x)(x+y)`

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

a: =(x+y)(x^2-xy+y^2)+(x+y)

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

b: =(x-y)(x^2+xy+y^2)+(x-y)

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

c: =x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2-y^3

=2x^3+6xy^2

d: =(x-y)^3+(y-x)(y+x)

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

Biến đổi tương đương nhé bạn.

a: Ta có: \(\left(x+y\right)^2\)

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

\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)

a: (x+y+z)^3-x^3-y^3-z^3

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

=(x+y)(y+z)(x+z)

b: x^3+y^3+z^3=1

x+y+z=1

=>x+y=1-z

x^3+y^3+z^3=1

=>(x+y)^3+z^3-3xy(x+y)=1

=>(1-z)^3+z^3-3xy(1-z)=1

=>1-3z-3z^2-z^3+z^3-3xy(1-z)=1

=>1-3z+3z^2-3xy(1-z)=1

=>-3z+3z^2-3xy(1-z)=0

=>-3z(1-z)-3xy(1-z)=0

=>(z-1)(z+xy)=0

=>z=1 và xy=0

=>z=1 và x=0; y=0

A=1+0+0=1

Bài 3:

a, (\(x\)+y+z)²

=((\(x\)+y) +z)²

= (\(x\) + y)² + 2(\(x\) + y)z + z²

= \(x^2\) + 2\(xy\) + y² + 2\(xz\) + 2yz + z²

=\(x^2\) + y² + z² + 2\(xy\) + 2\(xz\) + 2yz

b, (\(x-y\))(\(x^2\) + y² + z² - \(xy\) - yz - \(xz\))

= \(x^3\) + \(xy^2\) + \(xz^2\) - \(x^2\)y - \(xyz\) - \(x^2\)z - y³ 

Đến dây ta thấy xuất hiện \(x^3\) - y³ khác với đề bài, em xem lại đề bài nhé

c,

(\(x\) + y + z)³ 

=(\(x\) + y)³ + 3(\(x\) + y)²z + 3(\(x\)+y)z² + z³

= \(x^3\) + 3\(x^2\)y + 3\(xy^{2^{ }}\) + y³ +  3(\(x\)+y)z(\(x\) + y + z) + z³

= \(x^3\) + y³ + z³ + 3\(xy\)(\(x\) + y) + 3(\(x+y\))z(\(x+y+z\))

= \(x^3\) + y³ + z³ + 3(\(x\) + y)( \(xy\) + z\(x\) + yz + z²)

= \(x^3\) + y³ + z³ + 3(\(x\) + y){(\(xy+xz\)) + (yz + z²)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y){ \(x\)( y +z) + z(y+z)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y)(y+z)(\(x+z\)) (đpcm)