ĐK : 51x \(\ge0\Rightarrow x\ge0\)

Với \(x\ge0\)thì \(x+\frac{1}{1.3}>0;x+\frac{1}{3.5}>0;...;x+\frac{1}{99.101}>0\)

Khi đó : \(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{99.101}\right|=51x\)

<=> \(x+\frac{1}{1.3}+x+\frac{1}{3.5}+x+\frac{1}{5.7}+....+x+\frac{1}{99.101}=51x\)(50 hạng tử x ở VT)

<=> \(50x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=51x\)

<=> \(x=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\right)\)

<=> \(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

<=> \(x=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{50}{101}\)

Vậy x = 50/101 

=1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

=1-1/101

=100/101

k cho mình nha

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

TA CÓ \(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{99.101}\)

              \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)

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

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

 nhung ma ko cothoi gian giai

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

làm tắt thế ai mà hỉu đc

A=1 - 1/3+1/3 - 1/5+1/5 - 1/6+...+1/99 - 101+1/101 - 1/103

A=1 - 1/103

A=102/103

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{2}.\frac{102}{103}\)

\(=\frac{51}{103}\)

A = 1/1.3+1/3.5+...+1/99.101+1/101.103

 2A  = 2.(1/1.3+...+1/101.103)

        = 2.102/103

        => A= 102/103

\(\text{Đ}\text{ặt}:A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{99.101}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(A=\frac{100}{101}:2=\frac{50}{101}\)

\(\Rightarrow\frac{1}{3}x.x=\frac{50}{101}\)

\(x.\left(\frac{1}{3}.1\right)=\frac{50}{101}\)

\(x.\frac{1}{3}=\frac{50}{101}\)

$x=\frac{50}{101}:\frac{1}{3}=\frac{150}{101}$

\(.\frac{1}{3}x.x=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\frac{1}{3}xx=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(\frac{1}{3}xx=\frac{1}{2}.\left(\frac{100}{101}\right)\)

\(\frac{1}{3}xx=\frac{50}{101}\)

\(x.x=\frac{150}{101}\)

còn lại tự tính

\(\frac{1}{3}x.x=1-\frac{1}{101}=\frac{100}{101}\)

\(x.x=\frac{100}{101}:\frac{1}{3}=\frac{300}{101}\)

\(x=\sqrt{\frac{300}{101}}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(S=\frac{2}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(S=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(S=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

nhân S cho 2 

Công thức \(\frac{2}{x.\left(x+2\right)}=\frac{1}{x}-\frac{1}{x+2}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(S=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{99}-\frac{1}{101}\right).\frac{1}{2}\)

\(S=\left(\frac{1}{1}-\frac{1}{101}\right).\frac{1}{2}\)

\(S=\frac{100}{101}.\frac{1}{2}\)

\(S=\frac{50}{101}\)

a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

=2.(\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\))

=\(2.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=\(\frac{2}{2}.\left(1-\frac{1}{101}\right)\)

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

b, \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)

=\(5.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

=\(5.\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)

=\(\frac{250}{101}\)

\(=\frac{5}{2}.\frac{100}{101}\)

a,$\frac{2}{1.3}$+$\frac{2}{3.5}$+$\frac{2}{5.7}$+....+$\frac{2}{99.101}$

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

=>\(\frac{1}{1}-\frac{1}{101}\)

=>\(\frac{100}{101}\)

b,

$\frac{5}{1.3}$+$\frac{5}{3.5}$+$\frac{5}{5.7}$+....+$\frac{5}{99.101}$

a, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

=>\(\frac{1}{1}-\frac{1}{101}\)

=>\(\frac{100}{101}\)

b,\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+....+\frac{5}{99.101}\)

=>\(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{99.101}\right)\)

=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)

=>\(\frac{5}{2}.\frac{100}{101}\)

=>\(\frac{250}{101}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2S=1-\frac{1}{101}\Rightarrow2S+\frac{1}{101}=1\)