Answer: [tex]\dfrac{x^{2}+2x+1 }{\mathbf {x-1}} \cdot \dfrac{\mathbf {5x^{2} +15x-20} }{7x^{2} +7x}[/tex]
Step-by-step explanation:
[tex]\dfrac{\dfrac{5x^{2} +25x+20}{7x} }{\dfrac{x^{2} +2x+1}{7x^{2} +7x} } =\dfrac{5x^{2} +25x+20}{7x} \times \dfrac{7x^{2} +7x}{x^{2} +2x+1}=\dfrac{5x^{2} +25x+20}{7x} \times \dfrac{7x(x +1)}{(x+1)^{2} }=\\\\=\dfrac{5x^{2} +25x+20}{x+1} =\dfrac{{5(x+1)(x+4)}}{x+1}=5(x+4)[/tex]
[tex]5(x+4)=\dfrac{5(x-1) \cdot (x+4)}{x-1}=\dfrac{5x^{2} +15x-20}{x-1}[/tex]
Thus, the expressions will be used: (5x² + 15x - 20) and (x + 4).
Let's check:
[tex]\dfrac{x^{2}+2x+1 }{\mathbf {x-1}} \cdot \dfrac{\mathbf {5x^{2} +15x-20} }{7x^{2} +7x} =\dfrac{(x+1)^{2} }{x-1} \cdot \dfrac{5(x-1) \cdot (x+4)}{7x(x+1)} =\\\\=\dfrac{(x+1) \cdot 5 \cdot (x+4)}{7x} =\dfrac{5x^{2} +25x+20}{7x}[/tex]
