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\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{a\left(a+1\right)}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
1) \(\left[6.\left(-\frac{1}{3}\right)^3-3\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left[6.\frac{-1}{27}+1+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)
\(=\left[\frac{-2}{9}+2\right]:\frac{-4}{3}\)
\(=\left[\frac{-2}{9}+\frac{18}{9}\right]:\frac{-4}{3}\)
\(=\frac{16}{9}:\frac{-4}{3}\)
\(=\frac{-4}{3}.\)
2) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}.\)
Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x
\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)
\(\Rightarrow100x\ge0\)
\(\Rightarrow x\ge0\)
Từ điều kiện trên ta có :
\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)
\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(50x=1-\frac{1}{100}\)
\(50x=\frac{99}{100}\)
\(x=\frac{99}{5000}\)
Do \(\left|a\right|\ge0\forall a\) nên:
\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)
\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)
Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)
\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)
\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)
\(\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{1995\cdot1996}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{1995\cdot1996}\right)\)
\(=\left(1995\cdot1\right)-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1995}-\frac{1}{1996}\right)\)
\(=1995-\left(1-\frac{1}{1996}\right)\)
\(=1995-\frac{1995}{1996}\)
\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
\(D=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)
\(=2015-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=2015-\left(1-\frac{1}{2016}\right)\)
\(=2015-\frac{2015}{2016}\)
TO LẮM
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