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a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)
\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)
\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)
\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)
\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)
\(\frac{92}{17}:x=\frac{1}{17}\)
\(x=\frac{92}{17}:\frac{1}{17}\)
\(x=92\)
b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=1-\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
\(\Rightarrow x+3=19\)
\(\Rightarrow x=19-3\)
\(\Rightarrow x=16\)
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) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Rightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=11\\2x=-\frac{19}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{11}{2}\\x=-\frac{19}{4}\end{array}\right.\)
\(\left(1\frac{1}{4}-\frac{3}{5}\right):\frac{17}{20}< \frac{x}{17}< \left(5\frac{1}{3}-3\frac{1}{2}\right).\frac{12}{17}\)
= \(\left(\frac{5-3}{4}\right):\frac{17}{20}< \frac{x}{17}< \left(\frac{16}{3}-\frac{7}{2}\right).\frac{12}{17}\)
= \(\frac{1}{2}:\frac{17}{20}< \frac{x}{17}< \left(\frac{32-21}{6}\right).\frac{12}{17}\)
= \(\frac{10}{17}< \frac{x}{17}< \frac{3}{2}.\frac{12}{17}\)
= \(\frac{10}{17}< \frac{x}{17}< \frac{18}{17}\)
( Mik thấy mẫu giống nhau mik sẽ bỏ mẫu đi mik sẽ tìm tử )
=> 10 < 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 < 18
=> x = { 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 }
k mik nha làm ơn đó
Ta có : \(5\frac{8}{17}\div X+\left(-\frac{1}{17}\right)\div X+3\frac{1}{17}\div17\frac{1}{3}=\frac{4}{17}\)
Nên: \(\left(5\frac{8}{17}+\left(-\frac{1}{17}\right)\right)\div X+\frac{52}{17}\div\frac{52}{3}=\frac{4}{17}\)
\(5\frac{7}{17}\div X+\frac{52}{17}\times\frac{3}{52}=\frac{4}{17}\)
\(\frac{92}{17}\div X+\frac{3}{17}=\frac{4}{17}\)
\(\frac{92}{17}\div X=\frac{4}{17}-\frac{3}{17}\)
\(\frac{92}{17}\div X=\frac{1}{17}\)
\(X=\frac{92}{17}\div\frac{1}{17}\)
\(X=92\)
Vậy \(X=92\)
\(5\frac{8}{17}:x+\left(-\frac{1}{17}\right):x=\frac{52}{51}\)
\(\left(5\frac{8}{17}+-\frac{1}{17}\right):x=\frac{52}{51}\)
\(\frac{92}{17}:x=\frac{52}{51}\)
\(X=\frac{92}{17}:\frac{52}{51}=\frac{69}{13}\)