Wykonajmy obliczenia.

$1)2\frac{1}{3}+3\cdot \frac{2}{3}=2\frac{1}{3}+\frac{{}^1\cancel{3}}{1}\cdot \frac{2}{{\cancel{3}}^1}=2\frac{1}{3}+2=4\frac{1}{3}\to \text{B}$  

$2)\left(4,4-2\right)\cdot \frac{5}{8}=2,4\cdot \frac{5}{8}=2\frac{4}{10}\cdot \frac{5}{8}=2\frac{2}{5}\cdot \frac{5}{8}=\frac{{}^3\sout{12}}{{}_1\cancel{5}}\cdot \frac{{\cancel{5}}^1}{{\sout{8}}_2}=\frac{3}{2}=1\frac{1}{2}\to \text{A}$      

$3)\frac{4}{5}+2,4:0,4-0,2=\frac{8}{10}+24:4-0,2=0,8+6-0,2=6,6\to \text{N}$  

$4)10\frac{2}{3}-2\frac{1}{3}+\frac{2}{3}=8\frac{1}{3}+\frac{2}{3}=8\frac{3}{3}=9\to \text{A}$   

$5)3,6\cdot \frac{5}{12}-\frac{1}{2}\cdot 3=3\frac{6}{10}\cdot \frac{5}{12}-\frac{3}{2}=3\frac{3}{5}\cdot \frac{5}{12}-\frac{3}{2}=\frac{{}^3\sout{18}}{{}_1\cancel{5}}\cdot \frac{{\cancel{5}}^1}{{\sout{12}}_2}-\frac{3}{2}=\frac{3}{2}-\frac{3}{2}=0\to \text{C}$    

$6)\left(12-3,6:2\right)\cdot 4=\left(12-1,8\right)\cdot 4=10,2\cdot 4=40,8\to \text{H}$    

1	2	3	4	5	6
B	A	N	A	C	H

Stefan Banach