(\(\frac{1}{38}\)-1 ) (\(\frac{1}{37}\)-1 ) ( \(\frac{1}{36}\)-1 ) .......... ( \(\frac{1}{2}\)-1) = ?
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29 tháng 4 2020
\(\left(\frac{x+1}{39}+1\right)+\left(\frac{x+2}{38}+1\right)=\left(\frac{x+3}{37}+1\right)+\left(\frac{x+4}{36}+1\right)\)
\(\Leftrightarrow\frac{x+40}{39}+\frac{x+40}{38}-\frac{x+40}{37}-\frac{x+40}{36}=0\)
\(\Leftrightarrow\left(x+40\right)\left(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\right)=0\)
<=> x+40=0 (vì \(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\ne\)0)
<=> x=-40
Vậy x=-40
13 tháng 7 2019
\(\left(\frac{1}{38}-1\right).\left(\frac{1}{37}-1\right).\left(\frac{1}{36}-1\right)....\left(\frac{1}{2}-1\right)\)
\(=-\left(1-\frac{1}{38}\right).\left(1-\frac{1}{37}\right)....\left(1-\frac{1}{2}\right)\)
\(=-\frac{37}{38}.\frac{36}{37}...\frac{1}{2}\)
\(=\frac{-1}{38}\)
13 tháng 7 2019
\(\left(\frac{1}{38}-1\right)\left(\frac{1}{37}-1\right)\left(\frac{1}{36}-1\right).......\left(\frac{1}{2}-1\right)\)
\(=\frac{-37}{38}.\frac{-36}{37}.\frac{-35}{36}.......\frac{-2}{3}.\frac{-1}{2}\)
\(=\frac{-1}{38}\)
T
13 tháng 7 2019
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
TT
10 tháng 8 2018
\(\left(\dfrac{1}{38}-1\right).\left(\dfrac{1}{37}-1\right).\left(\dfrac{1}{36}-1\right)....\left(\dfrac{1}{2}-1\right)\)
\(=\left(\dfrac{1}{38}-\dfrac{38}{38}\right).\left(\dfrac{1}{37}-\dfrac{37}{37}\right).\left(\dfrac{1}{36}-\dfrac{36}{36}\right)....\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\)
\(=\left(\dfrac{-37}{38}\right).\left(\dfrac{-36}{37}\right).\left(\dfrac{-35}{36}\right)...\left(\dfrac{-1}{2}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(\dfrac{-1}{1}\right).\left(\dfrac{-1}{1}\right).....\left(-\dfrac{1}{1}\right)\)
\(=\left(\dfrac{-1}{38}\right).\left(-1\right).\left(-1\right).....\left(-1\right)\)
\(=\dfrac{\left(-1\right).\left(-1\right).....\left(-1\right)}{38}\)
\(=\dfrac{\left(-1\right)^{38}}{38}\)
\(=\dfrac{1}{38}\)
DN
16 tháng 8 2015
\(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}
8 tháng 9 2017
Ta có : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\)\(\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)
\(=\frac{1}{3}+\frac{3}{30}+\frac{3}{45}=\frac{1}{2}\)
\(\Rightarrow\)\(\frac{1}{3}+\frac{1}{31}+\frac{1}{36}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)
\(\RightarrowĐPCM\)
6 tháng 8 2016
\(\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{38.60}\)
\(=\frac{1}{2.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\frac{19}{20}=\frac{19}{120}\)
\(=\frac{-37}{38}\cdot\frac{-36}{37}\cdot...\cdot\frac{-1}{2}=\frac{\left(-37\right)\left(-36\right)\cdot...\cdot\left(-1\right)}{38\cdot37\cdot...\cdot2}=\frac{1}{38}\)