a: x-y-z=0
=>x=y+z; z=x-y; y=x-z
\(B=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
\(=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{z+y}{z}=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
b:Sửa đề: \(25-y^2=8\left(x-2022\right)^2\)
=>\(25-y^2>=0\)
=>\(y^2\le25\)
mà y nguyên
nên \(y^2\in\left\lbrace0;1;4;9;16;25\right\rbrace\)
Ta có: \(25-y^2=8\left(x-2022\right)^2\)
=>\(25-y^2\) ⋮8
mà \(y^2\in\left\lbrace0;1;4;9;16;25\right\rbrace\)
nên \(y^2\in\left\lbrace1;9;25\right\rbrace\)
TH1: \(y^2=1\)
Ta có: \(25-y^2=8\left(x-2022\right)^2\)
=>\(8\left(x-2022\right)^2=25-1=24\)
=>\(\left(x-2022\right)^2=3\) (vô lý vì x nguyên)
=>LOại
TH2: \(y^2=9\)
Ta có: \(25-y^2=8\left(x-2022\right)^2\)
=>\(8\left(x-2022\right)^2=25-9=16\)
=>\(\left(x-2022\right)^2=2\) (Vô lý vì x nguyên)
=>Loại
TH3: \(y^2=25\)
Ta có: \(25-y^2=8\left(x-2022\right)^2\)
=>\(8\left(x-2022\right)^2=25-25=0\)
=>\(\left(x-2022\right)^2=0\)
=>x-2022=0
=>x=2022
Ta có: \(y^2=25\)
=>\(\left[\begin{array}{l}y=5\\ y=-5\end{array}\right.\)
