Zad. 14
[tex]a) \\b-\frac{10b-5}{5}=b-\frac{5(2b-1)}5=b-(2b-1)=b-2b+1=-b+1\\b) \\\frac{8x-4}4-\frac{x}2=\frac{8x-4}4-\frac{2x}4=\frac{8x-4-2x}4=\frac{6x-4}4=\frac{2(3x-2)}4=\frac{3x-2}2\\c)\\2(n-3)-\frac{4n+1}2=\frac{4(n-3)}2-\frac{4n+1}2=\frac{4n-12}2-\frac{4n+1}2=\frac{4n-12-(4n+1)}2=\frac{4n-12-4n-1}2=\frac{-13}2=-\frac{13}2=-6\frac12=-6,5[/tex]
[tex]d) \\\frac{3a-4}5-2(6-5a)=\frac{3a-4}5-\frac{10(6-5a)}5=\frac{3a-4-(60-50a)}5=\frac{3a-4-60+50a}5=\frac{53a-64}5\\e) \\x-\frac{x-1}3+3(x-\frac19)=\frac{3x}3-\frac{x-1}3+3x-\frac13=\frac{3x-(x-1)}3+\frac{9x}3-\frac13=\frac{3x-(x-1)+9x-1}3=\frac{3x-x+1+9x-1}3=\frac{11x}3=\frac{11}3x[/tex]
[tex]f) \\\frac{2p+5}6+\frac{3-4p}3=\frac{2p+5}6+\frac{2(3-4p)}6=\frac{2p+5+2(3-4p)}6=\frac{2p+5+6-8p}6=\frac{-6p+11}6=-\frac{6p-11}6\\g) \\\frac{x+3}2-\frac{4x+2}4+\frac{x}3=\frac{6(x+3)}{12}-\frac{3(4x+2)}{12}+\frac{4x}{12}=\frac{6x+18-(12x+6)+4x}{12}=\frac{6x+18-12x-6+4x}{12}=\frac{-2x+12}{12}=-\frac{2x-12}{12}=-\frac{2(x-6)}{12}=-\frac{x-6}6[/tex]
[tex]h) \\2m+\frac{3m-1}5-\frac{2m-3}{15}=\frac{30m}{15}+\frac{3(3m-1)}{15}-\frac{2m-3}{15}=\frac{30m+9m-3-(2m-3)}{15}=\frac{39m-3-2m+3}{15}=\frac{37m}{15}\\\\i) \\\frac{2k-7}3-\frac{4+10k}6+\frac{k}5=\frac{10(2k-7)}{30}-\frac{5(4+10k)}{30}+\frac{6k}{30}=\frac{20k-70-(20+50k)+6k}{30}=\frac{20k-70-20-50k+6k}{30}=\frac{-24k-90}{30}=-\frac{24k+90}{30}=-\frac{6(4k+15)}{30}=-\frac{4k+15}5[/tex]
Zad. S4
[tex]\frac{5a(8a-6b)}2-(4a+b)(5a-2b)=\\\frac{40a^2-30ab}{2}-(20a^2-8ab+5ab-2b^2)=\\\frac{2(20a^2-15ab)}2-(20a^2-3ab-2b^2)=\\20a^2-15ab-20a^2+3ab+2b^2=\\2b^2-15ab+3ab=\\2b^2-12ab=\\2b(b-6a)[/tex]
Odp. C
