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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\) \(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\) hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\) Vậy: \(x=\frac{9}{10}\) b) Ta có: \(5\frac{4}{7}:x=13\) \(\Leftrightarrow\frac{39}{7}:x=13\) \(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\) Vậy: \(x=\frac{3}{7}\) c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\) \(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\) \(\Leftrightarrow x\cdot\frac{14}{5}=84\) \(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\) Vậy: x=30 d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\) \(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\) hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\) Vậy: x=-5 e) Ta có: \(8\frac{2}{3}:x-10=-8\) \(\Leftrightarrow\frac{26}{3}:x=2\) hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\) Vậy: \(x=\frac{13}{3}\) g) Ta có: \(x+30\%=-1.3\) \(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\) hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\) Vậy: \(x=\frac{-8}{5}\) i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\) \(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\) \(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\) \(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\) Vậy: x=-9 k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\) \(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\) \(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\) hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\) Vậy: x=30 m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\) \(\Leftrightarrow\left|2x-1\right|=16\) \(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\) Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\) g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\) \(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\) Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\) a) (x + 1/2) . (2/3 − 2x) = 0 \(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\) b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\) \(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-2+\frac{3}{7}\) \(\Rightarrow\left(x.\frac{44}{7}+\frac{3}{7}\right).\frac{11}{5}=-\frac{11}{7}\) \(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{11}{7}:\frac{11}{5}=-\frac{11}{7}.\frac{5}{11}\) \(\Rightarrow x.\frac{44}{7}+\frac{3}{7}=-\frac{5}{7}\) \(\Rightarrow x.\frac{44}{7}=-\frac{5}{7}-\frac{3}{7}\) \(\Rightarrow x.\frac{44}{7}=-\frac{8}{7}\) \(\Rightarrow x=-\frac{8}{7}:\frac{44}{7}=-\frac{8}{7}.\frac{7}{44}\) \(\Rightarrow x=-\frac{2}{11}\) c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\) \(\Rightarrow x\left(3\frac{1}{4}-\frac{7}{6}\right)=\frac{5}{12}+\frac{5}{3}\) \(\Rightarrow x\left(\frac{13}{4}-\frac{7}{6}\right)=\frac{25}{12}\) \(\Rightarrow x.\frac{25}{12}=\frac{25}{12}\) \(\Rightarrow x=\frac{25}{12}:\frac{25}{12}\) \(\Rightarrow x=1\) d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\) \(\Rightarrow\left(5\frac{8}{17}-\frac{4}{17}\right):x+\frac{22}{7}:\frac{52}{3}=\frac{4}{11}\) \(\Rightarrow5\frac{4}{17}:x+\frac{33}{182}=\frac{4}{11}\) \(\Rightarrow\frac{89}{17}:x=\frac{4}{11}-\frac{33}{182}\) \(\Rightarrow\frac{89}{17}:x=\frac{365}{2002}\) \(\Rightarrow x=\frac{89}{17}:\frac{365}{2002}\) \(\Rightarrow x\approx28,7\) (số hơi lẻ) e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\) \(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\) \(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\) \(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\) Giải: a) \(\frac{1}{5}-\frac{2}{3}+2x=\frac{1}{2}\) \(\Leftrightarrow2x=\frac{1}{2}-\left(\frac{1}{5}-\frac{2}{3}\right)\) \(\Leftrightarrow2x=\frac{1}{2}-\frac{-7}{15}\) \(\Leftrightarrow2x=\frac{11}{15}\) \(\Leftrightarrow x=\frac{11}{15}:2\) \(\Leftrightarrow x=\frac{11}{30}\) b) \(4\left(\frac{1}{3}-3\right)+\frac{1}{2}=\frac{5}{6}+x\) \(\Leftrightarrow\frac{-61}{6}=\frac{5}{6}+x\) \(\Leftrightarrow x=\frac{-61}{6}-\frac{5}{6}\) \(\Leftrightarrow x=\frac{-66}{6}=-11\) 1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\) \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\) \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\) \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\) \(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\) \(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\) \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\) \(\left(x-\frac{9}{4}\right)=\frac{45}{4}\) \(x=\frac{45}{4}+\frac{9}{4}\) \(x=\frac{27}{2}\)