Write as a single rational expression a) (1/x) - (1/(x-1))b) (a-b)/(a² + (1/a))c) (3x/2y) - (5x/6y) + (x/3y)

Answer:a) [tex]\frac{-1}{x^2-x)}[/tex]b) [tex]\frac{(a^2-ab)}{a^3+1}[/tex]c) [tex]\frac{7x}{6y}[/tex]Step-by-step explanation:Given:a) [tex]\frac{1}{x}-\frac{1}{x-1}[/tex]now,taking the LCM of the denominator, we havex(x-1)multiplying and dividing the given equation with the x(x-1)thus,[tex]\frac{x(x-1)}{x(x-1)}[\frac{1}{x}-\frac{1}{x-1}][/tex]or⇒ [tex]\frac{1}{x(x-1)}[\frac{1\times(x(x-1))}{x}-\frac{1\times(x(x-1))}{x-1}][/tex]or⇒ [tex]\frac{1}{x(x-1)}[\frac{(x-1)}{1}-\frac{x}{1}][/tex]or⇒ [tex]\frac{1}{x(x-1)}[x-1-x][/tex]or⇒ [tex]\frac{x-1-x}{x(x-1)}[/tex]or⇒ [tex]\frac{-1}{x(x-1)}[/tex]or⇒ [tex]\frac{-1}{x^2-x)}[/tex]b) [tex]\frac{(a-b)}{a^2+(\frac{1}{a})}[/tex]now,⇒ [tex]\frac{(a-b)}{(\frac{a^2\times a+1}{a})}[/tex]or⇒ [tex]\frac{(a-b)\times a}{a^3+1}[/tex]or⇒ [tex]\frac{(a^2-ab)}{a^3+1}[/tex]c) [tex]\frac{5x}{6y}+\frac{x}{3y}[/tex]taking LCM of the denominator, we have(6y × 3y)multiplying and dividing the given equation with the (6y × 3y)thus,⇒ [tex]\frac{6y\times3y}{6y\times3y}\frac{5x}{6y}+\frac{x}{3y}[/tex]or⇒ [tex]\frac{1}{6y\times3y}\frac{(6y\times3y)\times5x}{6y}+\frac{(6y\times3y)x}{3y}[/tex]or⇒ [tex]\frac{1}{6y\times3y}\frac{(3y)\times5x}{1}+\frac{(6y)x}{1}[/tex]or⇒ [tex]\frac{3y)\times5x+(6y)x}{6y\times3y}[/tex]or⇒ [tex]\frac{15xy+6xy}{18y^2}[/tex]or⇒ [tex]\frac{21xy}{18y^2}[/tex]or⇒ [tex]\frac{7x}{6y}[/tex]