\(\frac{tfrhttrty}{bbn,mnm}\)= 1\(\frac{8}{8}\)
hỏi t = ?
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\(\frac{x-1}{6}=\frac{1}{6}\Leftrightarrow6\left(x-1\right)=6\Leftrightarrow6x-6=6\Leftrightarrow6x=12\Leftrightarrow x=2\)
\(\frac{x-2}{5}=\frac{8}{10}=\frac{4}{5}\Leftrightarrow5\left(x-2\right)=4.5=20\Leftrightarrow5x-10=20\Leftrightarrow5x=30\Leftrightarrow x=6\)
\(\frac{x-1}{8}=\frac{1}{2}\Leftrightarrow2\left(x-1\right)=8\Leftrightarrow2x-2=8\Leftrightarrow2x=10\Leftrightarrow x=5\)
\(\text{ câu 1 mk nghĩ là so sánh chứ nhỉ?}\)
Tính :
a) \(\frac{1}{6}.\frac{9}{-8}.\left(-\frac{12}{11}\right)=\frac{1.9.\left(-12\right)}{6.\left(-8\right).11}=\frac{1.3.3}{2.2.11}=\frac{9}{44}\)
b) \(\frac{1}{8}.\frac{-3}{7}+\frac{1}{8}.\frac{-4}{7}\)
\(=\frac{1}{8}.\left(-\frac{3}{7}+\frac{-4}{7}\right)\)
\(=\frac{1}{8}.\frac{-7}{7}=\frac{1}{8}.\left(-1\right)\)
\(=-\frac{1}{8}\)
c) \(\frac{7}{13}.\frac{3}{-5}+\frac{6}{13}.\frac{-3}{5}\)
\(=-\frac{3}{5}.\left(\frac{7}{13}+\frac{6}{13}\right)\)
\(=-\frac{3}{5}.\frac{13}{13}=-\frac{3}{5}.1\)
\(=-\frac{3}{5}\)
a, \(\frac{1}{6}.\frac{9}{-8}.\frac{-12}{11}\)
\(\Rightarrow\frac{9}{8}.\frac{2}{11}\)
\(\Rightarrow\frac{9}{4}.\frac{1}{11}\)
\(\Rightarrow\frac{9}{44}\)
b,\(\frac{1}{8}.\frac{-3}{7}+\frac{1}{8}.\frac{-4}{7}\)
\(\Rightarrow\frac{-3}{56}-\frac{1}{2}.\frac{1}{7}\)
\(\Rightarrow\frac{-3}{56}-\frac{1}{2}.\frac{1}{7}\)
\(\Rightarrow\frac{-3}{56}-\frac{1}{14}\)
\(\Rightarrow\frac{-1}{8}\)
c,\(\frac{7}{13}.\frac{3}{-5}+\frac{6}{13}.\frac{-3}{5}\)
\(\Rightarrow\frac{-21}{65}-\frac{18}{65}\)
\(\Rightarrow\frac{-3}{5}\)
\(M=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\)
\(M=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(M=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(M=\frac{3}{-5}=\frac{-3}{5}\)
Ta có : \(M=\left[\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{18}-\frac{4}{9}\right)\right]:\left(\frac{494949}{525252}-1\right)\)
=> \(M=\left[\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}+\frac{1}{18}-\frac{4}{9}\right)\right]:\left(\frac{494949}{525252}-\frac{525252}{525252}\right)\)
=> \(M=\left(\frac{7}{8}:\frac{-2}{9}\right):\left(\frac{494949}{525252}-\frac{525252}{525252}\right)\)
=> \(M=\left(-\frac{63}{16}\right):\left(-\frac{3}{52}\right)\)
=> \(M=\frac{63}{16}:\frac{3}{52}=\frac{63}{16}.\frac{52}{3}=\frac{21.3.13.4}{4.4.3}=\frac{21.13}{4}=\frac{273}{4}\)
a, Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath
b, Câu hỏi của Vũ Xuân Hiếu - Toán lớp 6 | Học trực tuyến
c)
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{9}{18}-\frac{2}{18}\)
\(=\frac{7}{18}\)
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{7}{18}\)
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\(5\frac{3}{8}-\frac{40}{5}\left(x-1\right)=4\frac{3}{8}\)
\(\Leftrightarrow\frac{43}{8}-\frac{40}{5}.\left(x-1\right)=\frac{35}{8}\)
\(\Leftrightarrow\frac{40}{5}\left(x-1\right)=1\)
\(\Leftrightarrow40\left(x-1\right)=5\)
\(\Leftrightarrow40x=45\)
\(\Leftrightarrow x=\frac{9}{8}\)
Vậy \(x=\frac{9}{8}\)