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

\(a,=\left(x-y\right)\left(x+y\right)+11\left(x-y\right)=\left(x-y\right)\left(x+y+11\right)\\ b,=\left(x+z\right)\left(x^2-xz+z^2\right)+y\left(x^2+z^2-xz\right)\\ =\left(x^2-xz+z^2\right)\left(x+y+z\right)\)

a. x² - y² + 11x - 11y

= (x + y)(x - y) + 11(x - y)

= (x + y + 11)(x - y)

b. Mik ko hiểu đề lắm

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

1. Ta có: hằng đẳng thức: \(x^3+y^3+z^3=3xyz\) nếu x+y+z=0

đặt b-c=x, c-a=y, a-b=z⇒x+y+z=0

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

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

3. Tham khảo: https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-y-5-x-5-y-5-thanh-nhan-tu-faq447273.html

\(5,=x^3+2x^2y-7x^2y-14xy^2\\ =x^2\left(x+2y\right)-7xy\left(x+2y\right)\\ =x\left(x-7y\right)\left(x+2y\right)\)

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

Nếu a ≥ 0, b  ≥  0, c  ≥  0 thì :

Đề bài yêu cầu gì vậy bạn?

\(z^3\left(x+y^2\right)+y^3\left(z-x^2\right)-x^3\left(y+z^2\right)-xyz\left(xyz-1\right)\)

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

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

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

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

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

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

Tick hộ nha bạn 😘

z^3(x+y^2)+y^3(z-x^2)-x^3(y+z^2)-xyz(xyz-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)

\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

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

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2