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\(\left(\frac{1}{38}-1\right).\left(\frac{1}{37}-1\right).\left(\frac{1}{36}-1\right)....\left(\frac{1}{2}-1\right)\)
\(=-\left(1-\frac{1}{38}\right).\left(1-\frac{1}{37}\right)....\left(1-\frac{1}{2}\right)\)
\(=-\frac{37}{38}.\frac{36}{37}...\frac{1}{2}\)
\(=\frac{-1}{38}\)
\(\left(\frac{1}{38}-1\right)\left(\frac{1}{37}-1\right)\left(\frac{1}{36}-1\right).......\left(\frac{1}{2}-1\right)\)
\(=\frac{-37}{38}.\frac{-36}{37}.\frac{-35}{36}.......\frac{-2}{3}.\frac{-1}{2}\)
\(=\frac{-1}{38}\)
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(\left(\dfrac{1}{38}-1\right).\left(\dfrac{1}{37}-1\right).\left(\dfrac{1}{36}-1\right)....\left(\dfrac{1}{2}-1\right)\)
\(=\left(\dfrac{1}{38}-\dfrac{38}{38}\right).\left(\dfrac{1}{37}-\dfrac{37}{37}\right).\left(\dfrac{1}{36}-\dfrac{36}{36}\right)....\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\)
\(=\left(\dfrac{-37}{38}\right).\left(\dfrac{-36}{37}\right).\left(\dfrac{-35}{36}\right)...\left(\dfrac{-1}{2}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(\dfrac{-1}{1}\right).\left(\dfrac{-1}{1}\right).....\left(-\dfrac{1}{1}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(-1\right).\left(-1\right).....\left(-1\right)\)
\(=\dfrac{\left(-1\right).\left(-1\right).....\left(-1\right)}{38}\)
\(=\dfrac{\left(-1\right)^{38}}{38}\)
\(=\dfrac{1}{38}\)
\(=\frac{-37}{38}\cdot\frac{-36}{37}\cdot...\cdot\frac{-1}{2}=\frac{\left(-37\right)\left(-36\right)\cdot...\cdot\left(-1\right)}{38\cdot37\cdot...\cdot2}=\frac{1}{38}\)