Given:
The expressions are:
[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}[/tex]
[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}[/tex]
To find:
The simplified form of the given expressions.
Solution:
We have,
[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}[/tex]
It can be written as:
[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}=\dfrac{12a^{2+1}bc^3}{40b^4c^2}[/tex]
[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}=\dfrac{3a^{3}c^{3-2}}{10b^{4-1}}[/tex]
[tex]\dfrac{2a^2b}{2c^2}\cdot \dfrac{6ac^3}{20b^4}=\dfrac{3a^{3}c^{1}}{10b^3}[/tex]
Therefore, the value of the given expression is [tex]\dfrac{3a^{3}c}{10b^3}[/tex].
We have,
[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}[/tex]
It can be written as:
[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}=\dfrac{(x+y)(x-y)}{4(x+y)}\cdot \dfrac{x+y}{x-y}[/tex]
[tex]\dfrac{x^2-y^2}{4x+4y}\cdot \dfrac{x+y}{x-y}=\dfrac{x+y}{4}[/tex]
Therefore, the value of the given expression is [tex]\dfrac{x+y}{4}[/tex].
