Question

Помогите решить все во вложениях! 

63bed16f1c2638df1fe9adec8f34e5b5.docx

Answer 1

$1)\ 1frac{1}{6}+1frac{5}{6}*(1.854:1.8-1.5*2.02)=frac{7}{6}+frac{11}{6}*-2=frac{7-22}{6}=-frac{15}{6}\ \ \ 2)frac{z+2}{2z+4}+frac{2z+1}{3z-4}=frac{(z+2)(3z-4)+ (2z+1)(2z+4)}{(2z+4)(3z-4)}=frac{(z+2)(7z-2)}{2(z+2)(3z-4)}=frac{7z-2}{2(3z-4)}$

Answer 2

1_1/6 +1_5/6 * (1.854 :1.8 - 1.5 * 2.02)=
7/6+11/6 *( 1.03 - 3.03 )=
7/6+11/6 * (-2) =
7/6 - 22/6 =
- 15/6 = - 2.5

Упростите
(z - 2) /(2z+4) + (2z +1) /(3z -4)=
[(z - 2)(3z-4) +(2z+4)(2z +1)] / [(3z -4)(2z+4)]=
[3z^2 +8-6z-4z +4z^2+8z+2z +4] / [(3z -4)(2z+4)]=
(7z^2 +12 ) / [(3z -4)(2z+4)]