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a) \(\left( {\frac{7}{3} + 3,5} \right):\left( { - \frac{{25}}{6} + \frac{{22}}{7}} \right) + 0,5\)
\(\begin{array}{l} = \left( {\frac{7}{3} + \frac{7}{2}} \right):\left( { - \frac{{25}}{6} + \frac{{22}}{7}} \right) + \frac{1}{2}\\ = \frac{{35}}{6}:\frac{{ - 25.7 + 22.6}}{{6.7}} + \frac{1}{2}\\ = \frac{{35}}{6}:\frac{{ - 43}}{{7.6}} + \frac{1}{2} = \frac{{35}}{6}.\frac{{7.6}}{{ - 43}} + \frac{1}{2}\\ = \frac{{ - 245}}{{43}} + \frac{1}{2} = \frac{{ - 245.2 + 43}}{{43.2}} = \frac{{ - 447}}{{86}}\end{array}\)
b) \(\frac{{38}}{7} + \left( { - 3,25} \right) - \frac{{17}}{7} + 4,55\)
\(\begin{array}{l} = \left( {\frac{{38}}{7} - \frac{{17}}{7}} \right) + \left( {4,55 - 3,25} \right)\\ = \frac{{38 - 17}}{7} + 1,3 = \frac{{21}}{7} +1,3\\ = 3 + 1,3 = 4,3\end{array}\)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(\begin{array}{l}\left( {\frac{{20}}{7}.\frac{{ - 4}}{{ - 5}}} \right) + \left( {\frac{{20}}{7}.\frac{3}{{ - 5}}} \right) = \frac{{20}}{7}.\left( {\frac{{ - 4}}{{ - 5}} + \frac{3}{{ - 5}}} \right)\\ = \frac{{20}}{7}.\left( {\frac{{ - 1}}{{ - 5}}} \right) = \frac{{20}}{7}.\frac{1}{5} = \frac{{20}}{{35}} = \frac{4}{7}\end{array}\)
\(A=\frac{-173.24+42.\left(-173\right)-173.34}{-18.519+\left(-519\right).49+33.\left(-519\right)}=\frac{\left(-173\right).24+42.\left(-173\right)+\left(-173\right).34}{\left(-519\right).18+\left(-519\right).49+33.\left(-519\right)}\)
\(=\frac{\left(-173\right).\left(24+42+34\right)}{\left(-519\right).\left(18+49+33\right)}=\frac{\left(-173\right).100}{\left(-519\right).100}=\frac{-173}{-519}=\frac{1}{3}\)
\(B=\frac{10.11.12+20.22.24+30.33.36+40.44.48}{10.12.14+20.24.28+30.36.42+40.48.56}\)
\(=\frac{10.11.12+2.10.2.11.2.12+3.10.3.11.3.12+4.10.4.11.4.12}{10.12.14+2.10.2.12.2.14+3.10.3.12.3.14+4.10.4.12.4.14}\)
\(=\frac{10.11.12+8.10.11.12+27.10.11.12+64.10.11.12}{10.12.14+8.10.12.14+27.10.12.14+64.10.12.14}\)
\(=\frac{\left(10.11.12\right).\left(1+8+27+64\right)}{\left(10.12.14\right).\left(1+8+27+64\right)}=\frac{10.11.12}{10.12.14}=\frac{11}{14}\)