[tex]a)\frac{x^2+9x+20}{2x+8}\cdot\frac{4}{x^2-25}=\\
x^2+9x+20=x^2+4x+5x+20=x(x+4)+5(x+4)=\\
=(x+4)(x+5)\\
x^2-25=(x-5)(x+5)\\
Revenind:\frac{(x+4)(x+5)}{2(x+4)}\cdot\frac{4}{(x-5)(x+5)}=\boxed{\frac{2}{x-5}}\\
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b)\frac{x^2-10x+21}{2x^2-6x}\cdot\frac{4x^2-36}{x^2-14x+49}=\\
x^2-10x+21=x^2-3x-7x+21=x(x-3)-7(x-3)=\\
=(x-3)(x-7)\\
2x^2-6x=2x(x-3)\\
4x^2-36=4(x^2-9)=4(x-3)(x+3)\\
x^2-14x+49=(x-7)^2\\
Revenind:\frac{(x-3)(x-7)}{2x(x-3)}\cdot \frac{4(x-3)(x+3)}{(x-7)^2}=\frac{2(x-3)(x+3)}{x(x-7)}\\
[/tex]
[tex]c)\frac{x^2+10x+25}{x^2+6x+9}\cdot \frac{x^2-x-12}{x^2+2x-15}=\\
x^2+10x+25=(x+5)^2\\
x^2+6x+9=(x+3)^2\\
x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=\\
=(x+3)(x-4)\\
x^2+2x-15=x^2+5x-3x-15=x(x+5)-3(x+5)=\\
=(x+5)(x-3)\\
Revenind:\frac{(x+5)^2}{(x+3)^2}\cdot \frac{(x+3)(x-4)}{(x+5)(x-3)}=\frac{(x+5)(x-4)}{(x+3)(x-3)}\\
d)\frac{x^2-2x-8}{x^2+x-20}\cdot \frac{x^2+8x+15}{x^2+8x+12}=\\
x^2-2x-8=x^2-4x+2x-8=x(x-4)+2(x-4)=\\
=(x-4)(x+2)\\
x^2+x-20=x^2+5x-4x-20=x(x+5)-4(x+4)=\\
=(x+5)(x-4)\\
[/tex]
[tex]x^2+8x+15=x^2+3x+5x+15=x(x+3)+5(x+3)=\\
=(x+3)(x+5)\\
x^2+8x+12=x^2+6x+2x+12=x(x+6)+2(x+6)=\\
=(x+6)(x+2)\\
Revenind:\frac{(x-4)(x+2)}{(x+5)(x-4)}\cdot \frac{(x+3)(x+5)}{(x+6)(x+2)}=\frac{x+3}{x+6}
[/tex]