x 2 y + x y 2  +  x 2 z + x z 2  +  y 2 z + y z 2  + 3xyz.

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

= x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy)

= (x + y + z)(xy + xz + yz).

\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

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

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

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

Biến đổi tương đương nhé bạn.

a: Ta có: \(\left(x+y\right)^2\)

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

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

b) Thay x=-1; y=1 và z=-2 vào B, ta được:

\(B=\dfrac{3\cdot\left(-1\right)\cdot1\cdot\left(-2\right)-2\cdot\left(-2\right)^2}{\left(-1\right)^2+1}=\dfrac{6-8}{1+1}=\dfrac{-2}{2}=-1\)

chịu

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

\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)

\(C=ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b-a+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b\right)+bc\left(a-c\right)+ac\left(a-c\right)=b\left(a+b\right)\left(a-c\right)+c\left(a-c\right)\left(a+b\right)=\left(a+b\right)\left(c+c\right)\left(a-c\right)\)

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)