Cũng tương tự thôi c.

x³ + y³ + z³ - 3xyz = ( x³ + y³) + z³ - 3xyz

= ( x + y)³ - 3xy(x + y) + z³ - 3xyz = (x + y)³ + z³ - 3xy( x + y) - 3xyz

= (x + y)³ + z³ - 3xy(x + y + z)

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

= ( x + y + z )( x² + 2xy + y² - xz - yz + z² ) - 3xy( x + y + z )

= ( x + y + z )( x² + 2xy + y² - xz - yz + z² - 3xy )

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

Làm cho mk đi @Ender Dragon Boy Vcl

(x−y+z)2+(z−y)2+2(x−y+z)(y−z)(x−y+z)2+(z−y)2+2(x−y+z)(y−z)

=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)

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

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

=x2−xy+xz+(z−y)(−x)=x2−xy+xz+(z−y)(−x)

=x2−xy+xz−xz+xy=x2−xy+xz−xz+xy

=x2

đc chưa

đánh chữ mỏi tay :V

\(x;y;z\rightarrow q;h;p\)

\(=\left(q^2+h^2+p^2\right)\left(q^2+h^2+p^2+2qh+2hp+2qp\right)+\left(qh+hp+pq\right)^2\)

\(Dat:\hept{\begin{cases}q^2+h^2+p^2=f\\qh+hp+qp=g\end{cases}}\Rightarrow\left(p^2+h^2+q^2\right)\left(p+q+h\right)^2+\left(qh+pq+ph\right)^2\)

\(=f\left(f+2g\right)+g^2=f^2+2fg+g^2=\left(f+g\right)^2=\left(q^2+h^2+p^2+qh+hp+pq\right)^2\)

shitbo Cho đệ sửa lại bài SP chứ bài SP dài quá ạ:p

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

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

Đặt \(x^2+y^2+z^2=a;xy+yz+zx=b\)

\(\Rightarrow a\left(a+2b\right)+b^2=a^2+2ab+b^2=\left(a+b\right)^2=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)

Đặt \(x^4+y^4+z^4=a;x^2+y^2+z^2=b;x+y+z=c\)

Ta có:\(2a-b^2-2bc^2+c^4\)

\(=2a-2b^2+b^2-2bc^2+c^4\)

\(=2\left(a-b^2\right)+\left(b-c^2\right)^2\)

Lại có:

\(a-b^2=-2\left(x^2y^2+y^2z^2+z^2x^2\right);b-c^2=-2\left(xy+yz+zx\right)\)( Nhác quá hơi tắt xíu )

Thay vào ta được:

\(2\left(a-b^2\right)+\left(b-c^2\right)^2\)

\(=-4\left(x^2y^2+y^2z^2+z^2x^2\right)+4\left(x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)\right)\)

\(=4xyz\left(x+y+z\right)\)

Bài 2:

a)   Đặt:  x - y =a;   y - z = b;    z - x = c   thì   a + b + c = 0

C/M: đẳng thức phụ:   a³ + b³ + c³ = 3abc

Ta có: \(a+b+c=0\)

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

\(\Rightarrow\)\(\left(a+b\right)^3=-c^3\)

\(\Rightarrow\)\(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3=3abc\)

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

\(\left(x^2-6x\right)^2-2\left(x-3\right)^2-81=\left[\left(x^2-6x\right)^2-81\right]-2\left(x-3\right)^2=\left[\left(x^2-6x\right)^2-9^2\right]-2\left(x-3\right)^2=\left(x^2-6x+9\right)\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x+11\right)\)

=\(\left(x-3\right)^2\left(x^2-6x-11\right)\)

nha

`P=x^3/(x+y)+y^3/(y+z)+z^3/(z+x)`

`=x^4/(x^2+xy)+y^4/(y^2+yz)+z^4/(z^2+zx)`

Ad bđt cosi-swart:

`P>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+xy+yz+zx)`

Mà `xy+yz+zx<=x^2+y^2+z^2)`

`=>P>=(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2))=(x^2+y^2+z^2)/2=3/2`

Dấu "=" xảy ra khi `x=y=z=1`

`Q=(x^3+y^3)/(x+2y)+(y^3+z^3)/(y+2z)+(z^3+x^3)/(z+2x)`

`Q=(x^3/(x+2y)+y^3/(y+2z)+z^3/(z+2x))+(y^3/(x+2y)+z^3/(y+2z)+x^3/(z+2x))`

`Q=(x^4/(x^2+2xy)+y^4/(y^2+2yz)+z^4/(z^2+2zx))+(y^4/(xy+2y^2)+z^4/(yz+2z^4)+x^4/(xz+2x^2))`

Áp dụng BĐT cosi-swart ta có:

`Q>=(x^2+y^2+z^2)^2/(x^2+y^2+z^2+2xy+2yz+2zx)+(x^2+y^2+z^2)^2/(2(x^2+y^2+z^2)+xy+yz+zx))`

Mà`xy+yz+zx<=x^2+y^2+z^2`

`=>Q>=(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))+(x^2+y^2+z^2)^2/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2)^2)/(3(x^2+y^2+z^2))=(2(x^2+y^2+z^2))/3=2`

Dấu "=" xảy ra khi `x=y=z=1.`