Tính \(\frac{42}{1,4\times1,4}\)= ?
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a: =12h-7h=5h
b: =1,4h*1+1,4h+1,4h*8
=1.4h*10
=14h
c: =1,25h+1,25h+1,25h*8
=1,25h*10=12,5h
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{-2,6}{12}=\frac{x}{-42}\Leftrightarrow x=\frac{\left(-2,6\right)\left(-42\right)}{12}=9,1\)
\(\frac{1,4}{-5}=\frac{2,1}{x}\Leftrightarrow x=\frac{2,1.\left(-5\right)}{1,4}=-7,5\)
\(\frac{3}{x-1}=\frac{21}{16}\Leftrightarrow x-1=\frac{3.16}{21}=\frac{16}{7};x=\frac{23}{7}\)
\(\frac{1,2}{30}=\frac{3x+4}{50}\Leftrightarrow3x+4=\frac{50.1,2}{30}=2;3x=-1;x=-\frac{1}{3}\)
a) \(\frac{-2,6}{12}=\frac{x}{42}\)
\(x=\frac{-2,6.\left(-42\right)}{12}=9,1\)
b) \(\frac{1.4}{-5}=\frac{2,1}{x}\)
\(x=\frac{-5.2,1}{1,4}=-7,5\)
c) \(\frac{3}{x-1}=\frac{21}{16}\)
\(x-1=\frac{3.16}{21}=\frac{16}{7}\)
\(x=\frac{16}{7}+1=\frac{23}{7}\)
d) \(\frac{1,2}{30}=\frac{3x+4}{50}\)
\(3x+4=\frac{12.50}{30}=20\)
\(3x=20-4=16\)
\(x=16:3=\frac{16}{3}\)
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\(\begin{array}{l}a)0,25 + 1\frac{5}{{12}} = \frac{{25}}{{100}} + \frac{{17}}{{12}}\\ = \frac{1}{4} + \frac{{17}}{{12}} = \frac{3}{{12}} + \frac{{17}}{{12}}\\ = \frac{{20}}{{12}} = \frac{5}{3}\\b) - 1,4 - \frac{3}{5}\\ = \frac{{ - 14}}{{10}} - \frac{3}{5} = \frac{{ - 7}}{5} - \frac{3}{5}\\ = \frac{{ - 10}}{5} = - 2\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}-\frac{2}{7}\)
\(=0\)
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\frac{2}{7}-\frac{2}{7}=0\)
Study well ! >_<