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\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)
\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=-1+1+\frac{2}{97}\)
\(=\frac{2}{97}\)
\(a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4\)
\(=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}\)
\(b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)\)
\(=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}\)
\(=\frac{100.12}{123.16}+\frac{23.15}{123.20}\)
\(=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}\)
\(=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}\)
\(=\frac{123}{164}=\frac{3}{4}\)
\(=\left[\frac{3}{5}-\frac{1}{35}-\left(\frac{-3}{7}\right)\right]+\left[\frac{3}{11}-\frac{3}{4}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{21}{35}-\frac{1}{35}+\frac{15}{35}\right]+\left[\frac{12}{44}-\frac{33}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{20}{35}+\frac{15}{35}\right]+\left[\frac{-21}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=1+\left(-1\right)\)
\(=0\)
\(A=\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{6}-0,875+0,7}\)
\(=\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}\)
\(\left(-\frac{3}{4}+\frac{59}{9}\right):\left(-\frac{7}{23}\right)+\left(\frac{5}{9}-\frac{1}{4}\right):\frac{7}{23}\)
= \(\left(-\frac{3}{4}+\frac{59}{9}\right).\left(-\frac{23}{7}\right)+\left(\frac{5}{9}-\frac{1}{4}\right).\left(\frac{23}{7}\right)\)
=\(\left(-\frac{23}{7}\right).\left(-\frac{3}{4}+\frac{59}{9}-\frac{5}{9}+\frac{1}{4}\right)\)
=\(\left(-\frac{23}{7}\right).\left(-\frac{2}{4}+\frac{54}{9}\right)=\left(-\frac{23}{7}\right).\left(-\frac{1}{4}+6\right)\)
=\(\left(-\frac{23}{7}\right).\frac{11}{2}=-\frac{253}{14}\)
\(\frac{230^5}{23^5}=\frac{23^5.10}{23^5}=10\)
Quá hợp lí!
Học tốt nhá
230^5 sẽ có kết quả bằng 23^5 và 5 số 0 ở cuối
từ đó 230^5:23^5= 100000 (5 số 0)
#chanh