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

x 3  – x +  y 3  – y

= ( x 3  +  y 3 ) − (x + y)

= (x + y)( x 2  – xy + y 2 ) − (x + y)

= (x + y)(  x 2  – xy +  y 2  − 1)

= (√x - √y)(√x + √y)2

= (√x - √y)(√x + √y)(√x + √y)

= (x - y)(√x + √y)

x 3 - x + 3 x 2 y + 3 x y 2 + y 3 - y = x 3 + 3 x 2 y + 3 x y 2 + y 3 - x - y = x + y 3 - x - y = x + y x + y 2 - 1 = x + y x + y + 1 x + y - 1

a) 12x.                           b) 4xy

c) 2y(3 x 2   +   y 2 ).

d) (x + y + z)( x 2   +   y 2   + z 2  – xy – xz - yz).

a) ab + b√a + √a + 1 = [(√a)2b + b√a] + (√a + 1)

= b√a(√a + 1) + (√a + 1) = (√a + 1)(b√a + 1)

= (√x - √y)(√x + √y)2

= (√x - √y)(√x + √y)(√x + √y)

= (x - y)(√x + √y)

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

1.

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

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

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

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

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

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

= (x³ + 3x²y + 3xy² + y³) - (x+y)

= (x + y)³ - (x + y)

= (x + y).[(x+y)² - 1 ]

= (x + y).(x + y - 1).(x + y + 1)

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

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

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

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

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

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

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

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

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

#\(Urushi\text{☕}\)

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)

`x^3+y^3+x+y`

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

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