\(\left(x-y\right):\left(x+y\right):xy=1:7:24\)

\(\Rightarrow\frac{x-y}{1}=\frac{x+y}{7}=\frac{xy}{24}\) (1)

Áp dụng tính chất của dãy tỉ số bằng nhau đốt với hai tỉ số đầu ta có:

\(\frac{x-y}{1}=\frac{x+y}{7}=\frac{x-y+x+y}{1+7}=\frac{2x}{8}=\frac{x}{4}\)

Do đó \(\frac{x}{4}=\frac{xy}{24}\Rightarrow\frac{x}{xy}=\frac{4}{24}\Rightarrow\frac{1}{y}=\frac{1}{6}\Rightarrow y=6\)

Thay y = 6 vào (1) ta có:

\(\frac{x-6}{1}=\frac{x+6}{7}\)

=> 7(x - 6) = x + 6

=> 7x - 42 = x + 6

=> 7x - x = 6 + 42

=> 6x = 48

=> x = 8

Vậy x = 8, y = 6

Ta có: \(B=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)

Từ: \(x-y-z=0\Rightarrow x-z=y;y-x=-z\) và \(y+z=x\)

Suy ra: \(B=\frac{y}{x}.\frac{-z}{y}.\frac{x}{z}=-1\left(x;y;z\ne0\right)\)

x + y = 1

<=> (x + y)² = 1²

<=> x² + y² + 2xy = 1

<=> x² + y² = 1 - 2xy

Ta có:

\(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

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

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

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

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

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

\(=\dfrac{-2xy\left(x-y\right)}{xy\left(x^2y^2+3\right)}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

\(=-\dfrac{2\left(x-y\right)}{x^2y^2+3}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

= 0 (đpcm)