xem lại đề; x = 1 -> đề sai

Đề bài có lẽ bị sai , nếu thử x = 5 , y = 7 , z = 8 

Đề đâu có sai đâu Hoàng Lê Bảo Ngọc

\(\left(x^3+1\right)\left(y^3+1\right)\left(z^3+1\right)=\dfrac{81}{64}x^3y^3z^3\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{81}{64}x^2y^2z^2\)

\(\Leftrightarrow3xyz\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{81}{64}x^3y^3z^3\)

 \(\Rightarrow\left[{}\begin{matrix}xyz=0\\\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{27}{64}x^2y^2z^2\end{matrix}\right.\)

Nếu \(\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)=\dfrac{27}{64}x^2y^2z^2\) 

Ta có:

\(x^2-x+1=\dfrac{3}{4}x^2+\left(\dfrac{x}{2}-1\right)^2\ge\dfrac{3}{4}x^2\)

Tương tự: \(y^2-y+1\ge\dfrac{3}{4}y^2\) ; \(z^2-z+1\ge\dfrac{3}{4}z^2\)

Do các vế của các BĐT trên đều không âm, nhân vế với vế ta được:

\(\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge\dfrac{27}{64}x^2y^2z^2\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\) 

Thế vào  điều kiện \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=3xyz\) ko thỏa mãn (loại)

Vậy \(xyz=0\)

a)

\(x^3+y^3+z^3=x+y+z+2020\)

\(\Rightarrow\left(x^3-x\right)+\left(y^3-y\right)+\left(z^3-z\right)=2020\)

\(\Rightarrow x\left(x-1\right)\left(x+1\right)+y\left(y-1\right)\left(y+1\right)+z\left(z-1\right)\left(z+1\right)=2020\)

Ta có tích của ba số tự nhiên liên tiếp luôn chia hết cho 6

=> \(VT⋮6\)

Mà VP \(⋮̸\) 6

\(\Rightarrow\) Phương trình vô nghiệm

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

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

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

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

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

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

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

Ta có:

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

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

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

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

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

\(\Rightarrow\left(x+y+z\right)^3-\left(x+y+z\right)\left(3xy+3zx+3yz\right)=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{matrix}\right.\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

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

\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\)

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

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

\(\Rightarrow x=y=z\)

Xét trường hợp x = y = z, ta có:

\(P=\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(P=\dfrac{x^3}{2x.2x.2x}\)

\(P=\dfrac{x^3}{8x^3}\)

\(P=\dfrac{1}{8}\)

Xét trường hợp x + y + z = 0, ta có:

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

\(\Rightarrow P=\dfrac{-\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(\Rightarrow P=-1\)

Do x\(^3\)+y\(^3\)+z\(^3\)=3xyz\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

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

TH1:x+y+z=0\(\Rightarrow P=\frac{xyz}{\left(-z\right)\left(-y\right)\left(-x\right)}=-1\)

TH2:x=y=z\(\Rightarrow P=\frac{xyz}{8xyz}=\frac{1}{8}\)

a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

=0

c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{1}{xyz}\)

Ta có: x³ + y³ + z³ = 3xyz

x³ + y³ + z³ - 3xyz = 0

x³ + 3x²y + 3xy² + y³ + z³ - 3xy(x + y) - 3xyz = 0

(x + y)³ + z² - 3xy(x + y + z) = 0

(x + y + z)[(x + y)² - (x + y)z + z²] - 3xy(x + y + z) = 0

(x + y + z)(x² + 2xy + y² - xz - yz + z²) - 3xy(x + y + z) = 0

(x + y + z)(x² + 2xy + y² - xz - yz + z² - 3xy) = 0

(x + y + z)(x² + y² + z² - xz - yz - xy) = 0

=> x + y + z = 0 hoặc x² + y² + z² - xz - yz - xy = 0

+) Với x + y + z = 0 

<=> x + y = -z, x + z = -y, y + z = -x

Thay x + y = -z, x + z = -y, y + z = -x vào P, ta có:

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

+) Với x² + y² + z² - xz - yz - xy = 0

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

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

=> (x - y)² + (x - z)² + (y - z)² = 0

=> (x - y)² = 0 và (x - z)² = 0 và (y - z)² = 0

=> x = y và x = z và y = z

=> x = y = z

Thay x = y = z vào P, ta có:

\(P=\frac{xxx}{\left(x+x\right)\left(x+x\right)\left(x+x\right)}=\frac{x^3}{\left(2x\right)^3}=\frac{x^3}{8x^3}=\frac{1}{8}\)