\(\frac{-2}{5}x+\frac{2}{3}=\frac{7}{15}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
\(\frac{3}{7}\cdot15\cdot\frac{1}{3}+\frac{3}{7}\cdot5\cdot\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)
\(\Leftrightarrow\frac{15}{7}+\frac{6}{7}\le x\le-6\cdot\frac{-5}{3}\)
\(\Leftrightarrow3\le x\le10\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)
\(-\frac{2}{5}+\frac{5}{3}.\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{-7}{7
}\)
\(\frac{5}{3}.\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{-7}{6}-\left(-\frac{2}{5}\right)=-\frac{23}{30}\)
\(\left(\frac{3}{2}-\frac{4}{15}x\right)=-\frac{23}{30}:\frac{5}{3}=-\frac{23}{50}\)
\(\frac{4}{15}x=\frac{3}{2}-\left(-\frac{23}{50}\right)=\frac{49}{25
}\)
\(x=\frac{49}{25}:\frac{4}{15}\)
\(x=\frac{147}{20}\)
Vậy \(x\in\frac{147}{20}\)
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
\(-\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{7}{15}\)
=>\(-\dfrac{2}{5}x=\dfrac{7}{15}-\dfrac{2}{3}=\dfrac{7}{15}-\dfrac{10}{15}=-\dfrac{3}{15}=-\dfrac{1}{5}\)
=>\(x=\dfrac{-1}{5}:\dfrac{-2}{5}=\dfrac{1}{2}\)
\(\dfrac{-2}{5}x+\dfrac{2}{3}=\dfrac{7}{15}\\ \Rightarrow\dfrac{-2}{5}x=\dfrac{7}{15}-\dfrac{2}{3}\\ \Rightarrow\dfrac{-2}{5}x=\dfrac{-1}{5}\\ \Rightarrow x=\dfrac{-1}{5}:\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{-1}{5}.\dfrac{5}{-2}\\ \Rightarrow x=\dfrac{1}{2}\)
Vậy x = \(\dfrac{1}{2}\)