I need help with this please. :)
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Answer:
Step-by-step explanation:
1.
[tex]\displaystyle\frac{8x^2-4x}{4x-2}=\frac{4x(2x-1)}{2(2x-1)}=\boxed{2x}[/tex]
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2.
[tex]\displaystyle\frac{8x-4}{x+3}-\frac{12x-20}{2x+6}=\frac{8x-4}{x+3}-\frac{4(3x-5)}{2(x+3)}=\frac{8x-4-2(3x-5)}{x+3}\\\\=\frac{8x-4-6x+10}{x+3}=\frac{2x+6}{x+3}=\frac{2(x+3)}{x+3}=\boxed{2}[/tex]
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3.
[tex]\displaystyle\frac{(2x+10)^7}{(x+5)^6}=\frac{(2(x+5))^7}{(x+5)^6}=2^7(x+5)^{7-6}=128(x+5)\\\\=\boxed{128x+640}[/tex]