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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
\(=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)+\left(-\frac{3}{4}-\frac{2}{9}-\frac{1}{36}\right)+\frac{1}{64}\)
= 1 + -1 + 1/64
= 0 +1/64
= 1/64
\(A=5+\left|\frac{1}{3}-x\right|\)
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 \(x=\frac{1}{3}\)
\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)
=\(\frac{1}{99}-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\frac{97}{198}\)
=\(\frac{-95}{198}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\)
=>x=-1
vậy x=-1
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