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\(B=\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)...\left(\frac{2}{2008.2009}-1\right)\)
\(B=\left(\frac{2}{2.3}-\frac{6}{2.3}\right)\left(\frac{2}{3.4}-\frac{12}{3.4}\right)...\left(\frac{2}{2008.2009}-\frac{2008.2009}{2008.2009}\right)\)
\(B=\left(-\frac{4}{2.3}\right)\left(-\frac{10}{3.4}\right)...\left(\frac{2-2008.2009}{2008.2009}\right)\)
\(B=\left(-\frac{1.4}{2.3}\right)\left(-\frac{2.5}{3.4}\right)...\left(-\frac{2007.2010}{2008.2009}\right)\)
Biểu thức B có (2008 - 2) : 1 + 1 = 2007 (thừa số)
Vì cả 2007 thừa số của biểu thức B đều mang dấu (-)
Nên biểu thức B mang dấu (-)
\(B=-\frac{1.2....2007}{2.3...2008}.\frac{4.5...2010}{3.4...2009}\)
\(B=-\frac{1}{2008}.\frac{2010}{3}\)
\(B=-\frac{1.2010}{2008.3}=-\frac{1.1005}{1004.3}=-\frac{1.335}{1004.1}\)
\(B=-\frac{335}{1004}\)
Vậy\(B=-\frac{335}{1004}\)
\(\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{1995\cdot1996}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{1995\cdot1996}\right)\)
\(=\left(1995\cdot1\right)-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1995}-\frac{1}{1996}\right)\)
\(=1995-\left(1-\frac{1}{1996}\right)\)
\(=1995-\frac{1995}{1996}\)
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)
\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)
\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{a.\left(a+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{a}-\frac{1}{a+1}\)
\(=1-\frac{1}{a+1}\)
b, \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{a.\left(a+1\right).\left(a+2\right)}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{a.\left(a+1\right)}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
Chúc bạn học giỏi nha!!!
K cho mik vs nhé Hang Nguyen
Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900
Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi
\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)
\(=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)
\(=-4-\frac{1}{2}\)
\(=-\frac{9}{2}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(A=\left(3+\frac{1}{2}-\frac{2}{3}\right)-\left(2-\frac{2}{3}+\frac{5}{2}\right)+\left(-5+\frac{5}{2}-\frac{4}{3}\right)\)
\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)
\(A=\left(3-2-5\right)+\left(\frac{1}{2}-\frac{5}{2}+\frac{5}{2}\right)-\left(\frac{2}{3}-\frac{2}{3}+\frac{4}{3}\right)\)
\(A=-4+\frac{1}{2}-\frac{4}{3}\)
\(A=-\frac{29}{6}\)
Ta có : \(A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+.......+\left(1-\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=\left(1+1+1+......+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2016.2017}\right)\)
\(\Rightarrow A=2016-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(\Rightarrow A=2016-\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=2016-\frac{2016}{2017}=2015\frac{1}{2017}\)