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]

Your answer should be A

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