x 3  +  y 3  +  z 3  – 3xyz = x + y 3  – 3xy(x + y) +  z 3  – 3xyz

      = [  x + y 3  +  z 3 ] - [ 3xy.(x+ y) + 3xyz]

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

      = (x + y + z)[ x + y 2  – (x + y)z +  z 2 ] – 3xy(x + y + z)

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

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

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

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

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

b: a+b+c<>0

A=(a+b+c)^3-a^3-b^3-c^3/a+b+c

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

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

=1/2[a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2]

=1/2[(a-b)^2+(b-c)^2+(a-c)^2]>=0

\(x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Leftrightarrow x^2+y^2+z^2-xy-xz-yz=0\)

\(\Leftrightarrow x=y=z\)

Ta rút gọn tử thức trc: \(x^3+y^3+z^3-3xyz=x^3+y^3+z^3+x^2y-x^2y+xy^2-xy^2+y^2z-y^2z+yz^2-yz^2+x^2z-x^2z+xz^2-xz^2-xyz-xyz-xyz=x^2\left(x+y+z\right)+y^2\left(x+y+z\right)+z^2\left(x+y+z\right)-x\left(x+y+z\right)-yz\left(x+y+z\right)-xz\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=\frac{1}{2}\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\right)=\frac{1}{2}\left(x+y+z\right)\left(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right)\)tới đây rút gọn đc rồi chứ

\(a,=\left(2x-5\right)\left(x+1\right)\\ b,=\left(x-10\right)\left(x+1\right)\\ c,=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

Vế trái bằng vế phải nên đẳng thức được chứng minh.

Nếu x ≥ 0, y  ≥  0, z  ≥  0 thì:

x + y + z  ≥  0

x - y 2 + y - z 2 + z - x 2 ≥ 0

Suy ra:

x 3 + y 3 + z 3 - 3 x y z ≥ 0 ⇔ x 3 + y 3 + z 3 ≥ 3 x y z

Hay:  x 3 + y 3 + z 3 3 ≥ x y z

Lời giải:

\(A=\frac{x^3-y^3-z^3-3xyz}{(x+y)^2+(y-z)^2+(x+z)^2}=\frac{(x-y)^3+3xy(x-y)-z^3-3xyz}{x^2+y^2+2xy+y^2-2yz+z^2+z^2+x^2+2xz}\)

\(=\frac{(x-y)^3-z^3+3xy(x-y-z)}{2x^2+2y^2+2z^2+2xy-2yz+2xz}=\frac{(x-y-z)[(x-y)^2+z(x-y)+z^2]+3xy(x-y-z)}{2(x^2+y^2+xy-yz+xz)}\)

\(=\frac{(x-y-z)[(x-y)^2+z(x-y)+z^2+3xy]}{2(x^2+y^2+xy-yz+xz)}=\frac{(x-y-z)(x^2+y^2+z^2+xy-yz+xz)}{2(x^2+y^2+z^2+xy-yz+xz)}=\frac{x-y-z}{2}\)

Ta có: \(x^3+y^3+z^3-3xyz\)

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

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

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

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

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

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

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)

a) 16(12 t 2  +1).

b) Gợi ý x 3   +   y 3 = ( x   +   y ) 3  - 3xy(x + y)

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