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\(\frac{3}{2}+\frac{3}{14}+\frac{3}{15}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{6}{\left(1.2\right).2}+\frac{6}{\left(2.7\right).2}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(x-3\right).x}=\frac{96}{49.2}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{96}{98}\)
=> \(1-\frac{1}{x}=\frac{48}{49}\)
=> \(\frac{1}{x}=\frac{1}{49}\)
=> \(x=49\)
a. 60%x + 0,4x + x : 3 = 2
0.6x + 0,4x + x : 3 = 2
x(0,6 + 0,4 : 3 ) = 2
\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)
câu B tự làm nha .
\(\left(\frac{x}{-5}+1\frac{1}{2}\right):\frac{28}{75}-1,4.\frac{15}{49}=\left|-\frac{2}{3}\right|.\left(-\frac{3}{2}\right)^3\)
\(\left(\frac{x}{-5}+\frac{3}{2}\right).\frac{75}{28}-\frac{14}{10}.\frac{15}{49}=\frac{2}{3}.\frac{-27}{8}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}-\frac{3}{7}=\frac{-9}{4}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-9}{4}+\frac{3}{7}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-63}{28}+\frac{12}{28}\)
\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-51}{28}\)
\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}:\frac{75}{28}\)
\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}.\frac{28}{75}\)
\(\frac{-x}{5}+\frac{3}{2}=\frac{-17}{25}\)
\(\frac{-x}{5}=\frac{-17}{25}-\frac{3}{2}\)
\(\frac{-x}{5}=\frac{-34}{50}-\frac{75}{50}\)
\(\frac{-x}{5}=\frac{-109}{50}\)
\(\frac{-10x}{50}=\frac{-109}{50}\)
Hình như đề sai thì phải
a)\(\frac{5}{6}-x=-\frac{7}{12}+\frac{2}{3}\)
\(\frac{5}{6}-x=\frac{1}{12}\)
\(x=\frac{5}{6}-\frac{1}{12}\)
\(\Rightarrow x=\frac{3}{4}\)
b)\(\left(2,4x-36\right):1\frac{5}{7}=-14\)
\(\left(2,4x-36\right)=-24\)
\(2,4x=12\)
\(\Rightarrow x=5\)
c)\(\left(3\frac{1}{2}+2x\right).3\frac{2}{3}=5\frac{1}{3}\)
\(3\frac{1}{2}+2x=\frac{16}{11}\)
\(2x=-\frac{45}{22}\)
\(x=-\frac{45}{44}\)
d)\(\frac{5}{6}-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{3}{8}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{11}{24}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{11}{24}\\\frac{1}{2}x-\frac{1}{3}=-\frac{11}{24}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{19}{12}\\x=-\frac{1}{4}\end{cases}}\)
e)\(\left|\frac{1}{4}-2x\right|-\frac{3}{4}=0\)
\(\left|\frac{1}{4}-2x\right|=\frac{3}{4}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-2x=\frac{3}{4}\\\frac{1}{4}-2x=-\frac{3}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{2}\end{cases}}\)
