a)
[tex]\dfrac{4-5x}2+x=\dfrac{4-5x}2+\dfrac{2x}2=\dfrac{4-5x+2x}2=\boxed{\dfrac{4-3x}2}[/tex]
b)
[tex]2x-\dfrac{7x+1}4=\dfrac{8x}{4}-\dfrac{7x+1}4=\dfrac{8x-(7x+1)}{4}=\dfrac{8x-7x-1}{4}=\boxed{\dfrac{x-1}4}[/tex]
c)
[tex]\dfrac{2x+1}6+\dfrac{1-x}3=\dfrac{2x+1}{6}+\dfrac{2(1-x)}{6}=\dfrac{2x+1+2-2x}6=\dfrac36=\boxed{\dfrac12}[/tex]
d)
[tex]\dfrac{4x-5}3-\dfrac{x+4}2=\dfrac{2(4x-5)}{6}-\dfrac{3(x+4)}{6}=\dfrac{8x-10-(3x+12)}{6}=\dfrac{8x-10-3x-12}{6}=\\\\=\boxed{\dfrac{5x-22}{6}}[/tex]
e)
[tex]\dfrac{6x-12y}3+x-4y=\dfrac{3(2x-4y)}3+x-4y=2x-4y+x-4y=\boxed{3x-8y}[/tex]
f)
[tex]\dfrac{15a+20b}5-\dfrac{4a+8b}2=\dfrac{5(3a+4b)}5-\dfrac{2(2a+4b)}2=3a+4b-(2a+4b)=\\\\=3a+4b-2a-4b=\boxed{a}[/tex]
g)
[tex]\dfrac{7k+21}{14}-\dfrac{3k-9}6=\dfrac{7(k+3)}{14}-\dfrac{3(k-3)}6=\dfrac{k+3}2-\dfrac{k-3}2=\dfrac{k+3-(k-3)}2=\\\\=\dfrac{5+3-k+3}{2}=\dfrac62=\boxed{3}[/tex]
h)
[tex]4\!\!\!\!\diagup^1\cdot\dfrac{3p+6}{12\!\!\!\!\diagup_3}+\dfrac{10-5p}{15\!\!\!\!\diagup_5}\cdot9\!\!\!\!\diagup^3=\dfrac{3(p+2)}3+\dfrac{5(2-p)\cdot3}5=p+2+3(2-p)=\\=p+2+6-3p=\boxed{8-2p}[/tex]
