Question

номер 734 помогите пожалусто​

Answer 1

Ответ:

1)19/18х

2)-1/4z

3)-13/16y

4)-136/15a

Answer 2

Пошаговое объяснение:

1) $\frac{2}{3}x-\frac{1}{6}x+\frac{8}{9}x=\frac{2*6}{3*6}x-\frac{1*3}{6*3}x+\frac{8*2}{9*2}x=\frac{12}{18}x-\frac{3}{18}x+\frac{16}{18}x=\frac{12-3+16}{18}x=\frac{25}{18}x=1\frac{7}{18}x$

2) $\frac{23}{24}z-\frac{5}{8}z-\frac{7}{12}z=\frac{23}{24}z-\frac{5*3}{8*3}z-\frac{7*2}{12*2}z=\frac{23}{24}z-\frac{15}{24}z-\frac{14}{24}z=\frac{23-15-14}{24}z=-\frac{6}{24}z=-\frac{1}{4}z$

3) $-\frac{15}{16}y+\frac{3}{8}y-\frac{1}{4}=-\frac{15}{16}y+\frac{3*2}{8*2}y-\frac{1*4}{4*4}=-\frac{15}{16}y+\frac{6}{16}y-\frac{4}{16}=\frac{-15+6-4}{16}y=-\frac{13}{16}y$

4) $8\frac{1}{15}a+3\frac{2}{3}a-20\frac{4}{5}a=\frac{8*15+1}{15}a+\frac{3*3+2}{3}a-\frac{20*5+4}{5}a=\frac{121}{15}a+\frac{11}{3}a-\frac{104}{5}a=\frac{121}{15}a+\frac{11*5}{3*5}a-\frac{104*3}{5*3}a=\frac{121}{15}a+\frac{55}{15}a-\frac{312}{15}a=\frac{121+55-312}{15}a=-\frac{136}{15}a=-9\frac{1}{15}a$