Respuesta :
Answer:
The product [tex]\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}[/tex]
Step-by-step explanation:
Given expression [tex]\frac{b-5}{2b}[/tex] and [tex]\frac{b^2+3b}{b-5}[/tex]
We have to find the product of [tex]\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}[/tex]
Consider the given expression [tex]\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}[/tex]
Multiply fractions, we have,
[tex]\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}[/tex]
[tex]=\frac{\left(b-5\right)\left(b^2+3b\right)}{2b\left(b-5\right)}[/tex]
Cancel common factor ( b - 5 )
we have, [tex]=\frac{b^2+3b}{2b}[/tex]
Apply exponent rule,
[tex]\:a^{b+c}=a^ba^c[/tex]
[tex]b^2=bb[/tex]
[tex]=bb+3b=b(b+3)[/tex]
[tex]=\frac{b\left(b+3\right)}{2b}[/tex]
Cancel common factor b , we have,
[tex]=\frac{b+3}{2}[/tex]
Thus, the product [tex]\frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}[/tex]