(3/5)x²-[(4/5)x²+(2/3)x]+[1·(2/5)x²+0,(6)x]-0,2x²=
3x²/5-(4x²/5+2x/3)+(2x²/5+6x/9)-2x²/10=
3x²/5-4x²/5+2x²/5-2x²/10-2x/3+6x/9=
x²/5-2x²/10-2x/3+2x/3=
x²/5-x²/5=0
[0,(7)x-0,(6)a]+[0,(6)x-0,(5)a]-[(20/45)x-(10/45)a]=
7x/9-6a/9+6/9x-5a/9-4x/9+2a/9=
7x/9+6x/9-4x/9-6a/9-5a/9+2a/9=
9x/9-9a/9=x-a
[2,(3)a-1,(3)b]+[0,(6)a-0,(6)b]-(3a-2b)=
[(23-2)a/9]-[13-1b)/9]+6a/9-6b/9-3a+2b=
21a/9-12b/9+2a/3-2b/3-3a+2b=
7a/3+2a/3-3a-4b/3-2b/3+2b=
9a/3-3a-6b/3+2b=
3a-3a-2b+2b=0