\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)

Ta có \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\) (đpcm)

Áp dụng công thức trên ta có

A\(=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot\cdot\cdot\cdot\cdot+\frac{1}{2015\cdot2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2015\cdot2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{2}{3\cdot4}+....+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\)

\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\)

\(\Rightarrow A=\left(\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right)\div2\approx0.25\)

Vậy A\(\approx0.25\)

a) \(\left|2x-1\right|=5\)

\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\left[\begin{matrix}=3\\=-2\end{matrix}\right.\)

b) \(\left(5^x-1\right)3-2=70\)

\(\Rightarrow5^x.3-3=72\)

\(\Rightarrow5^x.3=75\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

c) \(\left(x-1\frac{1}{2}\right)^2+\frac{3}{4}=\frac{1}{4}\)

\(\Rightarrow\left(x-1\frac{1}{2}\right)^2=\frac{-1}{2}\)

............. Làm tiếp nhé!

d) \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

a)Xét 1/2-1/3-1/6=3/6-2/6-1/6=0

=> (1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).(1/2-1/3-1/6)=(1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).0=0

b) 4A=1.2.3.4+2.3.4.4+..+x(x+1)(x+2)4

         =1.2.3.4+2.3.4.5-1.2.3.4+...+x(x+1)(x+2)(x+3)-x(x+1)(x+2)(x-1)

         = (x-1)x(x+1)(x+2)

=> A=x(x+1)(x+2)(x-1)/4

1)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)

Sửa đề chút nha

\(\frac{x}{2}=\frac{1}{1.2.3}+....+\frac{1}{98.99.100}\)

Ta có công thức tổng quát  \(\frac{1}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{2}\left(\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\right)\)

\(\Rightarrow\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)

Áp dụng vào tổng ta có

\(\frac{x}{2}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{99.100}=\frac{4949}{9900}\)

\(\Rightarrow x=\frac{4949}{4950}\)

                                   Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

                                      \(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\) 

                                    \(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

                                 \(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

                                \(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

                                \(A=\frac{1}{2}.\left(\frac{4950-1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)

                             Vậy \(A=\frac{4949}{19800}\)

                                      Ủng hộ mk nha các bn !!!

A = \(\frac{4949}{19800}\)

A = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + ...+ \(\frac{1}{98.99.100}\)

2A = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{98.99.100}\)

2A = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{99}\).(\(\frac{2}{98.100}\))

2A = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{99}\).(\(\frac{1}{98}-\frac{1}{100}\))

2A = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{98.99}\) - \(\frac{1}{99.100}\)

2A = \(\frac12-\frac{1}{9900}\)

2A = \(\frac{4949}{9900}\)

A = \(\frac{4949}{9900}\) : 2

A = 4949/19800