Tính
\(\frac{45}{19}-\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^{-1}\right)^{-1}\)
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\(\frac{\left(1+17\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right)...\left(1+\frac{17}{19}\right)}{\left(1+19\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right)...\left(1+\frac{19}{17}\right)}\)
\(=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{18.19.20...36}{1.2.3...19}:\frac{20.21.22...36}{1.2.3...17}\)
\(=\frac{18.19.20...36}{1.2.3...19}.\frac{1.2.3...17}{20.21.22....36}=\frac{1.2.3...17.18...36}{1.2.3...19.20...36}=1\)
\(M=\frac{18.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{20.\frac{21}{2}.\frac{22}{3}...\frac{36}{17}}=\frac{\frac{18.19.20...36}{2.3...19}}{\frac{20.21.22...36}{2.3...17}}=\frac{\frac{18.19}{18.19}}{1}=\frac{1}{1}=1\)
Dễ thôi :)
\(\frac{45}{19}-\left\{\frac{1}{2}+\left[\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right]^{-1}\right\}^{-1}\)
\(=\frac{45}{19}-\left\{\frac{1}{2}+\left[\frac{1}{3}+4\right]^{-1}\right\}^{-1}\)
'\(=\frac{45}{19}-\left\{\frac{1}{2}+\frac{3}{13}\right\}^{-1}\)
\(=\frac{45}{19}-\frac{26}{19}\)
\(=\frac{19}{19}=1\)
=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