177. Ҳисоб кунед: а) $(5\frac{3}{7}-3\frac{2}{5})+(3-1\frac{17}{35})-(4\frac{12}{15}-3\frac{19}{105})$; б) $(7\frac{7}{9}-5\frac{1}{3})-(2-1\frac{5}{6})+(9\frac{19}{36}-3\frac{11}{12})$.

Question

Вопрос: $Фото задания: 177. Ҳисоб кунед: а) $(5\frac{3}{7}-3\frac{2}{5})+(3-1\frac{17}{35})-(4\frac{12}{15}-3\frac{19}{105})$; б)…$ Фото задания

Accepted Answer

а)

Аввал ҳар як қавсро ҳисоб мекунем:

$5\frac{3}{7}-3\frac{2}{5}=\frac{38}{7}-\frac{17}{5}=\frac{190-119}{35}=\frac{71}{35}=2\frac{1}{35}$.

$3-1\frac{17}{35}=\frac{105}{35}-\frac{52}{35}=\frac{53}{35}=1\frac{18}{35}$.

$4\frac{12}{15}-3\frac{19}{105}=4\frac{84}{105}-3\frac{19}{105}=1\frac{65}{105}=1\frac{13}{21}$.

Акнун натиҷаҳоро ҷамъу тарҳ мекунем:

$\frac{71}{35}+\frac{53}{35}-\frac{65}{105}=\frac{213+159-65}{105}=\frac{307}{105}=2\frac{97}{105}$.

б)

Ҳар як қавсро ҳисоб мекунем:

$7\frac{7}{9}-5\frac{1}{3}=\frac{70}{9}-\frac{16}{3}=\frac{70-48}{9}=\frac{22}{9}$.

$2-1\frac{5}{6}=\frac{12}{6}-\frac{11}{6}=\frac{1}{6}$.

$9\frac{19}{36}-3\frac{11}{12}=9\frac{19}{36}-3\frac{33}{36}=5\frac{22}{36}=5\frac{11}{18}$.

Натиҷаҳоро бо махраҷи умумии 18 ҷамъ мекунем:

$\frac{22}{9}-\frac{1}{6}+5\frac{11}{18}=\frac{44}{18}-\frac{3}{18}+\frac{101}{18}=\frac{142}{18}=\frac{71}{9}=7\frac{8}{9}$.

Ҷавоб: а) $2\frac{97}{105}$; б) $7\frac{8}{9}$.

177. Ҳисоб кунед: а) \((5\frac{3}{7}-3\frac{2}{5})+(3-1\frac{17}{35})-(4\frac{12}{15}-3\frac{19}{105})\); б) \((7\frac{7}{9}-5\frac{1}{3})-(2-1\frac{5}{6})+(9\frac{19}{36}-3\frac{11}{12})\).