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\(\frac{45}{19}-\left\{\frac{1}{2}+\left[\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right]^{-1}\right\}^{-1}\)
\(=\frac{45}{19}-\left\{\frac{1}{2}+\left[\frac{1}{3}+4\right]^{-1}\right\}^{-1}\)
'\(=\frac{45}{19}-\left\{\frac{1}{2}+\frac{3}{13}\right\}^{-1}\)
\(=\frac{45}{19}-\frac{26}{19}\)
\(=\frac{19}{19}=1\)
=45/19-(1/2+(1/3+4)^-1)^-1
=45/19-(1/2+(13/3)^-1)^-1
=45/19-(1/2+3/13)^-1
=45/19-(19/26)^-1
=45/19-26/19=19/19=1
câu hỏi hay......nhưng tui xin nhường cho các bn khác
Hãy tích đúng cho tui nha
THANKS
\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)...\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right)}\)
\(=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{18.19.20...36}{1.2.3...19}:\frac{20.21.22...36}{1.2.3...17}\)
\(=\frac{18.19.20...36}{1.2.3...19}.\frac{1.2.3...17}{20.21.22....36}=\frac{1.2.3...17.18...36}{1.2.3...19.20...36}=1\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)