помогите пожалуйста, заранее спасибо). Created by darinamusinarin. matematika-ru

помогите пожалуйста, заранее спасибо)...

NNNLLL54

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$1)\ \displaystyle \ 15\frac{3}{8}:\Big(2\frac{3}{4}\, x+5\frac{5}{6}\Big)-1\frac{1}{2}=\frac{3}{4}\\\\\\\frac{123}{8}:\Big(\frac{11}{4}\, x+\frac{35}{6}\Big)-\frac{3}{2}=\frac{3}{4}\\\\\\\frac{123}{8}:\frac{33x+70}{12}=\frac{3}{4}+\frac{3}{2}\\\\\\\frac{123\cdot 12}{8\cdot (33x+70)}=\frac{3+6}{4}\\\\\\\frac{123\cdot 3}{2\cdot (33x+70)}=\frac{9}{4}\ \ ,\ \ \ 33x+70=\frac{123\cdot 3\cdot 4}{2\cdot 9}\ \ ,\ \ \ 33x+70=82\ \ ,\\\\\\33x=12\ \ ,\ \ \ x=\frac{12}{33}\ \ ,\ \ \ x=\frac{4}{11}$

$2)\ \ \displaystyle 3\frac{1}{3}-\Big(4\frac{1}{5}\, x+x\Big):5\frac{4}{7}=\frac{8}{15}\\\\\\\frac{10}{3}-\frac{26}{5}\, x:\frac{39}{7}=\frac{8}{15}\\\\\\\frac{10}{3}-\frac{26x\cdot 7}{5\cdot 39}=\frac{8}{15}\ \ ,\ \ \ \frac{26x\cdot 7}{5\cdot 39}=\frac{10}{3}-\frac{8}{15}\ \ ,\ \ \ \frac{26x\cdot 7}{5\cdot 39}=\frac{50-8}{15}\ \ ,\\\\\\x=\frac{42\cdot 5\cdot 39}{15\cdot 7\cdot 26}\ \ ,\ \ \ x=3$

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Iryn95

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Пошаговое объяснение:

$\displaystyle 1) 15\frac{3}{8}:(2\frac{3}{4}x+5\frac{5}{6})-1\frac{1}{2}=\frac{3}{4}\\ \\ 15\frac{3}{8}:(2\frac{3}{4}x+5\frac{5}{6})=\frac{3}{4}+1\frac{1}{2}\\ \\ \frac{123}{8}:(2\frac{3}{4}x+5\frac{5}{6})=\frac{3}{4}+\frac{3}{2}\\ \\ \frac{123}{8}:(2\frac{3}{4}x+5\frac{5}{6})=\frac{3}{4}+\frac{6}{4}\\ \\ \frac{123}{8}:(2\frac{3}{4}x+5\frac{5}{6})=\frac{9}{4} \\ \\ (2\frac{3}{4}x+5\frac{5}{6})=\frac{123}{8}:\frac{9}{4} \\ \\ (2\frac{3}{4}x+5\frac{5}{6})=\frac{123}{8}*\frac{4}{9}$

$\displaystyle \frac{11}{4}x+\frac{35}{6}=\frac{123}{18}\\ \\\frac{11}{4}x=\frac{123}{18}-\frac{35}{6}\\ \\ \frac{11}{4}x=\frac{123}{18}-\frac{35*3}{18}\\ \\ \frac{11}{4}x=\frac{123}{18}-\frac{105}{18}\\ \\ \frac{11}{4}x=\frac{18}{18}\\ \\ \frac{11}{4}x=1\\ \\ x=1:\frac{11}{4}\\ \\ x=\frac{4}{11}$

$\displaystyle 2)3\frac{1}{3}-(4\frac{1}{5}x+x):5\frac{4}{7}=\frac{8}{15}\\ \\-(4\frac{1}{5}x+x):5\frac{4}{7}=\frac{8}{15}-\frac{10}{3}\\ \\ (4\frac{1}{5}x+x):5\frac{4}{7}=\frac{8-50}{15}\\ \\ (4\frac{1}{5}x+x):5\frac{4}{7}=-\frac{42}{15} \\ \\(4\frac{1}{5}x+x):5\frac{4}{7}=-\frac{14}{5}\\ \\ (4\frac{1}{5}x+x)=-\frac{14}{5}*\frac{39}{7} \\ \\ -5\frac{1}{5}x =-\frac{78}{5}\\ \\ x=-\frac{78}{5}:(-\frac{26}{5})=-\frac{78}{5}*(-\frac{5}{26)}= 3$

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