b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

A=1-1/2+1/2-1/3+1/3-1/4+...+1/64-1/128

A=1-1/128

A=127/128

Vậy A=\(\frac{127}{128}\)

B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

B=1-1/100

B=99/100

Vậy B=\(\frac{99}{100}\)

a=1/1x2+1/2x3+....+1/99x100

a=1-1/2+1/2-1/3+....+1/99-1/100

a=1-1/100

a=99/100

b=4/1x3+4/3x5+.....+4/51x53

b=2x(2/1x3+2/3x5+....+2/51x53)

b=2x(1-1/3+1/3-1/5+...+1/51-1/53)

b=2x(1-1/53)

b=2x52/53

b=104/53

đúng tick cho mình nha

Bài này cũng dễ mà

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(A=1+0-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}< 1\)

\(\Rightarrow A< 1\)

A=1⋅21​+2⋅31​+3⋅41​+....+99⋅1001​

�=1−12+12−13+13−14+...+199−1100A=1−21​+21​−31​+31​−41​+...+991​−1001​

�=1+(−12+12−13+13−14+...+199)−1100A=1+(−21​+21​−31​+31​−41​+...+991​)−1001​

�=1+0−1100A=1+0−1001​

�=1−1100<1A=1−1001​<1

$\Rightarrow A<1$

A<1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}\)

=\(\frac{99}{100}\)

A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100

A = 1 - 1/100

A = 99/100

A=1/1x2+1/2x3+...+1/99x100

A=1-1/2+1/2-1/3+1/3-...+1/99-1/00

A=1-1/100

A=99/100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100=99/100

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

= 99/100

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{100\cdot101}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)