Задание 509

Решение:

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

• \({\largeб)}\ \frac{1}{x-y}-\frac{1}{y-x}=\frac{1}{x-y}+\frac{1}{x-y}=\frac{1+1}{x-y}=\frac{2}{x-y}\)

• \({\largeв)}\ \frac{2a}{a-b}-\frac{3a}{b-a}=\frac{2a}{a-b}+\frac{3a}{a-b}=\frac{2a+3a}{a-b}=\frac{5a}{a-b}\)

• \({\largeг)}\ \frac{4m-1}{n-m}+\frac{m-4}{m-n}=\frac{4m-1}{n-m}-\frac{m-4}{n-m}=\frac{ 4m-1-(m-4)}{n-m}=\frac{4m-1-m+4}{n-m}=\frac{3m+3}{n-m}=\frac{ 3(m+1)}{n-m}\)

• \({\largeд)}\ \frac{2p+q}{p-2q}+\frac{p+3q}{2q-p}=\frac{2p+q}{p-2q}-\frac{p+3q}{p-2q}=\frac{ 2p+q-(p+3q)}{p-2q}=\frac{2p+q-p-3q}{p-2q}=\frac{p-2q}{p-2q}=1\)

• \({\largeе)}\ \frac{8a+b}{1-a}-\frac{2a-3b}{a-1}=\frac{8a+b}{1-a}+\frac{2a-3b}{1-a}=\frac{8a+b+2a-3b}{1-a}=\frac{10a-2b}{1-a}=\frac{ 2(5a-b)}{1-a}\)