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ta có \(\frac{1}{13}+\frac{16}{7}+\frac{3}{105}-\frac{9}{7}-\left(-\frac{12}{13}\right)\)
\(=\left[\frac{1}{13}-\left(-\frac{12}{13}\right)\right]+\left(\frac{16}{7}-\frac{9}{7}\right)+\frac{3}{105}\)
\(=1+1+\frac{3}{105}\)
\(=2+\frac{3}{105}=2\frac{3}{105}\)
a) \(\frac{7}{15}+\frac{9}{10}+\frac{8}{15}-\frac{-1}{10}-\frac{20}{10}+\frac{1}{157}\)
\(=\frac{7}{15}+\frac{9}{10}+\frac{8}{15}+\frac{1}{10}-\frac{20}{10}+\frac{1}{157}\)
\(=\left(\frac{7}{15}+\frac{8}{15}\right)+\left(\frac{9}{10}+\frac{1}{10}\right)-2+\frac{1}{157}\)
\(=1+1-2+\frac{1}{157}\)
\(=2-2+\frac{1}{157}\)
\(=0+\frac{1}{157}=\frac{1}{157}\)
b) \(\frac{1}{13}+\frac{16}{7}+\frac{3}{105}-\frac{9}{7}-\frac{-12}{13}\)
\(=\frac{1}{13}+\frac{16}{7}+\frac{1}{35}-\frac{9}{7}+\frac{12}{13}\)
\(=\left(\frac{1}{13}+\frac{12}{13}\right)+\left(\frac{16}{7}-\frac{9}{7}\right)+\frac{1}{35}\)
\(=1+1+\frac{1}{35}\)
\(=2+\frac{1}{35}\)
\(=\frac{70}{35}+\frac{1}{35}=\frac{71}{35}\)
\(\frac{210}{207}+\frac{105}{113}-\frac{3}{207}+\frac{8}{113}+27\)
\(=\left(\frac{210}{207}-\frac{3}{207}\right)+\left(\frac{105}{113}+\frac{8}{113}\right)+27\)
\(=1+1+27\)
\(=29\)
\(\frac{210}{207}+\frac{105}{113}-\frac{3}{207}+\frac{8}{113}+27\)
\(=\left(\frac{210}{207}-\frac{3}{2017}\right)+\left(\frac{105}{113}+\frac{8}{113}\right)+27\)
\(=\frac{207}{207}+\frac{113}{113}+27\)
\(=1+1+27\)
\(=29\)
Ta có: 1-5+9-13+...+105-109+113
=(1+9+...+105+113) -(5+13+...+109)
=(1+9+...+105+113)-(1+9+...+105+113)+14*8+1
=113