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

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

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

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

\(\Leftrightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

\(\Leftrightarrow x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(B=\dfrac{16.\left(-z\right)}{z}+\dfrac{3.\left(-x\right)}{x}-\dfrac{2019.\left(-y\right)}{y}=2019-19=2000\)

GIÁO VIÊN SAO TOÀN SAI HẰNG ĐẲNG THỨC THẾ????

Ta có: \(x^3+y^3+z^3=3xyz\)

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

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

   \(\Leftrightarrow\left(x+y+z\right).\left[\left(x+y+z\right)^2-3.\left(x+y\right).z\right]-3xy\left(x+y+z\right)=0\)

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

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

+ \(x+y+z=0\)\(\Rightarrow\)\(C=\frac{x^{2019}+y^{2019}+z^{2019}}{0}\)( Loại )

+ \(x^2+y^2+z^2-xz-yz-xy=0\)

\(\Rightarrow2x^2+2y^2+2z^2-2xz-2yz-2xy=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\)\(x=y=z\)

\(\Rightarrow\)\(C=\frac{x^{2019}+x^{2019}+x^{2019}}{\left(x+x+x\right)^{2019}}=\frac{3.x^{2019}}{3^{2019}.x^{2019}}=\frac{1}{3^{2018}}\)

Vậy.......

Từ x³ + y³ + z³ = 3xyz

=> ( x + y + z )( x² + y² + z² - xy - yz - xz ) = 0 ( phân tích như bạn kia )

Vì x + y + z ≠ 0

=> x² + y² + z² - xy - yz - xz = 0

<=> 2x² + 2y² + 2z² - 2xy - 2yz - 2xz = 0

<=> ( x - y )² + ( y - z )² + ( x - z )² = 0

VT ≥ 0 ∀ x,y,z. Đẳng thức xảy ra <=> x=y=z

Khi đó \(C=\frac{x^{2019}+y^{2019}+z^{2019}}{\left(x+y+z\right)^{2019}}=\frac{3x^{2019}}{\left(3x\right)^{2019}}=\frac{3x^{2019}}{3^{2019}\cdot x^{2019}}=\frac{1}{3^{2018}}\)

Bài 2:

a, \(x^3+y^3+z^3-3xyz\)

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

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

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

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

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

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

2a ) Ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)

2

a

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)

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

b

Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)

Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)

Áp dụng kết quả câu a ta được:

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

Sửa đề: x+y+z=0

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

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

=>\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

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

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

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

=>\(\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]=0\)(1)

x<>y<>z

=>\(x-y< >0;y-z< >0;x-z< >0\)

=>\(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ne0\left(2\right)\)

Từ (1),(2) suy ra x+y+z=0

Để chứng minh C/M: a+b+c=0, ta sẽ sử dụng công thức Newton về tổng hệ số của đa thức. Theo công thức Newton, ta có: (x+y+z)^3 = x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz Áp dụng công thức này vào phương trình đã cho, ta có: 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz = 3xyz Simplifying the equation, we get: x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2 + 2xyz = 0 Từ đây, ta có thể nhận thấy rằng: (x+y+z)(xy+yz+zx) = 0 Vì a, b, c là các số thực và a ≠ b ≠ c, nên ta có thể kết luận rằng: xy + yz + zx = 0 Do đó, ta có: (x+y+z)(xy+yz+zx) = (x+y+z)(0) = 0 Vì vậy, ta có C/M: a+b+c=0.... 

Từ \(a+b+c=0\Rightarrow a+b=-c\)

Xét hiệu \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)

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

  Thay \(a+b=-c;a+b+c=0\left(GT\right)v\text{ào}\left(I\right)\) ta được 

\(a^3+b^3+c^3-3abc=\left(-c\right)^3+c^3-3ab.0\)

                                         \(=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\left(\text{Đ}PCM\right)\)

Vậy \(a^3+b^3+c^3=3abc\) với \(a+c+b=0\)

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

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

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

\(x^2+y^2+z^2-xy-xz-yz=0\left(x+y+z\ne0\right)\)

\(2\times\left(x^2+y^2+z^2-xy-xz-yz\right)=0\times2\)

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

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

\(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)

\(\left[\begin{array}{nghiempt}x-y=0\\x-z=0\\y-z=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=y\\x=z\\y=z\end{array}\right.\)

x = y = z

\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{x}{z}\right)\)

\(=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)\)

\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)

\(=2^3\)

\(=8\)

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

(x+y+z) ( (x+y)^2 +z^2 -z(x+y) -3xy) =0

(x+y+z) ( x^2+ 2xy+y^2 +z^2- zx-zy-3xy)=0

(x+y+z) ( x^2+y^2+z^2 -zx-zy -xy)=0

Suy ra x+y+z =0 

x+y = -z

y+z = -x

x+z = -y

B = -16 + (-3) +2038 = 2019

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

\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\left(x,y,z\ne0\right)\)

+) x + y + z = 0 \(\Rightarrow B=\frac{-16z}{z}+\frac{-3x}{x}-\frac{-2038y}{y}\)

\(=-16-3+2038=2019\)

+) x = y = z \(\Rightarrow B=\frac{16.2z}{z}+\frac{3.2x}{x}-\frac{2038.2y}{y}\)

\(=32+6-4076=-4038\)

Cho em hỏi chút ạ, trường hợp  x=y=z suy ra ntn ạ?