$a)2x\left(3x+5\right)+3\left(x^2-3x+2\right)=$

$=\underline{6x^2}+\underline{\underline{10x}}+\underline{3x^2}-\underline{\underline{9x}}+6=$  

$=9x^2+x+6$  

$b)3\left(2x^2+x-7\right)-2x\left(x-3\right)=$  

$=\underline{6x^2}+\underline{\underline{3x}}-21-\underline{2x^2}+\underline{\underline{6x}}=$

$=4x^2+9x-21$  

 

$c)\frac{3}{4}\left(2\frac{2}{3}x-12\right)+\frac{2}{3}x\left(6x-4\frac{1}{2}\right)=$

$=\frac{3}{4}\left(\frac{8}{3}x-12\right)+\frac{2}{3}x\left(6x-\frac{9}{2}\right)=$

$=\frac{{\sout{3}}^1}{{\sout{4}}^1}\cdot \frac{{\sout{8}}^2}{{\sout{3}}^1}x-\frac{3}{{\sout{4}}^1}\cdot {\sout{12}}^3+\frac{2}{{\sout{3}}^1}x\cdot {\sout{6}}^{2}x-\frac{{\sout{2}}^1}{{\sout{3}}^1}x\cdot \frac{{\sout{9}}^3}{{\sout{2}}^1}=$

$=2x-9+4x^2-3x=4x^2-x-9$

 

$d)\frac{1}{3}\left(1\frac{1}{2}x-2\frac{1}{4}\right)-\frac{1}{2}\left(3x-\frac{1}{2}\right)=$

$=\frac{1}{3}\left(\frac{3}{2}x-\frac{9}{4}\right)-\frac{1}{2}\left(3x-\frac{1}{2}\right)=$

$=\frac{1}{2}x-\frac{3}{4}-\frac{3}{2}x+\frac{1}{4}=$

$=-\frac{2}{2}x-\frac{2}{4}=-x-\frac{1}{2}$