Question

[lethua]

giúp mình nhé

Answer 1

[Nguyễn Minh Quang]

bài 1

a.\(A=\frac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}+\frac{\left(x-x^2+1\right)\left(x+x^2-1\right)}{\left(x^2+x-1\right)\left(x^2+x+1\right)}+\frac{\left(x^2-x+1\right)\left(x^2-x-1\right)}{\left(x^2-x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{x^2+x-1}{x^2+x+1}+\frac{x-x^2+1}{x^2+x+1}+\frac{x^2-x+1}{x^2+x+1}=\frac{x^2+x+1}{x^2+x+1}=1\)

b. ta có \(\frac{x}{xy+x+1}=\frac{x}{xy+x+xyz}=\frac{1}{y+1+yz}=\frac{xyz}{y+xyz+yz}=\frac{xz}{xz+z+1}\)

tương tự ta sẽ có 

\(B=\frac{x}{xy+x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+xyz}=\frac{1}{1+y+yz}+\frac{1}{yz+y+1}+\frac{1}{xz+z+1}\)

\(=\frac{xy}{xy+x+1}+\frac{yz}{yz+y+1}+\frac{xz}{xz+z+1}\)

cộng lại ta có : \(3B=\frac{xy+x+1}{xy+x+1}+\frac{yz+y+1}{yz+y+1}+\frac{xz+z+1}{xz+z+1}=3\Rightarrow B=1\)