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\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+4}{2001}=\frac{x+4}{2002}+\frac{x+4}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\Rightarrow x+2004=0\)

=>x=-2004

vậy x=-2004

tham khảo bài của mình tại  http://olm.vn/hoi-dap/question/133172.html

16 tháng 7 2015

\(=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)+\left(-\frac{3}{4}-\frac{2}{9}-\frac{1}{36}\right)+\frac{1}{64}\)

= 1 + -1 + 1/64 

= 0 +1/64 

= 1/64

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\)

=>x=-1

vậy x=-1

24 tháng 6 2017

\(A=5+\left|\frac{1}{3}-x\right|\)

          Vì \(\left|\frac{1}{3}-x\right|\ge0\Rightarrow\left|\frac{1}{3}-x\right|+5\ge5\)

        Dấu = xảy ra khi \(\frac{1}{3}-x=0\Rightarrow x=\frac{1}{3}\)

Vậy Min A = 5 khi \(x=\frac{1}{3}\)

16 tháng 7 2015

\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)

=\(\frac{1}{99}-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\frac{97}{198}\)

=\(\frac{-95}{198}\)

19 tháng 8 2018

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

<=>  \(\frac{x+4}{2000}+1+\frac{x+3}{2001}=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)

<=>  \(\frac{x+4}{2000}+\frac{x+4}{2001}=\frac{x+4}{2002}+\frac{x+4}{2003}\)

<=>  \(\left(x+4\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

<=>  \(x+4=0\)   do 1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0

<=>  \(x=-4\)

Vậy...