\(\left(1-\frac{1}{97}\right)x\left(1-\frac{1}{98}\right)x...x\left(1-\frac{1}{1000}\right)\)
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12 tháng 3 2020
a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)
\(\Rightarrow x=-100\)
b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)
\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)
\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)
12 tháng 3 2020
b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0
\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0
\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))
\(\Leftrightarrow\)x=-1999
Vậy x=-1999
10 tháng 4 2016
(1 - 1/97) x (1 - 1/98) x ... x (1 - 1/1000)
= 96/97 x 97/98 x ... x 999/1000
= 12/125
MN
26 tháng 1 2016
\(A=\left(1-\frac{1}{97}\right)x\left(1-\frac{1}{98}\right)x....x\left(1-\frac{1}{1000}\right)\)
\(A=\frac{96}{97}x\frac{97}{98}x..x\frac{999}{1000}\)
\(A=\frac{96x97x98x...x999}{97x98x99x...x1000}=\frac{96}{1000}=\frac{12}{125}\)
HT
3 tháng 6 2016
Ta có \(\left(1-\frac{1}{97}\right)\times\left(1-\frac{1}{98}\right)\times.....\times\left(1-\frac{1}{1000}\right).\)
\(=\frac{97-1}{97}\times\frac{98-1}{98}\times.....\times\frac{1000-1}{1000}\)
\(=\frac{96}{97}\times\frac{97}{98}\times....\times\frac{999}{1000}\) (rút gọn hết )
\(=\frac{96}{1000}\)
\(=\frac{12}{125}\)
2 tháng 3 2020
\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)
\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))
\(⇔x=23\)
\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)
\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)
\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))
\(⇔x=-100\)
\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)
\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)
\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)
\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))
\(⇔x=-2013\)
KS
2 tháng 3 2020
\(\frac{201-x}{99}+\frac{203}{97}=\frac{205}{95}+3\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
Giúp mk với ạ
\(\left(1-\frac{1}{97}\right)x\left(1-\frac{1}{98}\right)x...x\left(1-\frac{1}{1000}\right)\)
\(\frac{96}{97}\cdot x\cdot\frac{97}{98}\cdot x\cdot...\cdot x\cdot\frac{999}{1000}\)
\(\frac{96}{97}\cdot\frac{97}{98}\cdot...\cdot\frac{999}{1000}\cdot x^{903}\)
\(\frac{96}{1000}\cdot x^{903}\)
\(\frac{12}{125}\cdot x^{903}\)
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