Respuesta :
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 have
x(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]
