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

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

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

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

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

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

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

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

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

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

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

nhân S cho 2 

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

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

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

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

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

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

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

\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

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

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

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

\(\frac{1}{1.3}+\frac{1}{3,5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

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

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\right)\)

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

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

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

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

-.-

1/1.3+1/3.5+1/5.7+...+1/99.101

=1/2.(2/1.3+2/3.5+2/5.7+...+2/99.101

=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)

=1/2.(1-1/101)

=1/2.100/101

=50/101

X+(1/1.3+1/3.5+1/5.7+...+1/99.101)=100

X+(2/1.3+2/3.5+2/5.7+...+2/99.101)=100

X+(1 -1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)=100

X+(1-1/101)=100

X+100/101=100

X=100-100/101

X=10000/101

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

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)

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

\(A=1-\frac{1}{100}=\frac{99}{100}\)

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

\(B=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)

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

\(B=1-\frac{1}{101}=\frac{100}{101}\)

\(C=\frac{3^2}{10}+\frac{3^2}{40}+\frac{3^2}{88}+...+\frac{3^2}{340}\)

\(C=3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\right)\)

\(C=3\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\right)\)

\(C=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(C=3\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{27}{20}\)

\(D=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)

\(D=\frac{7}{2}B=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)

\(E=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3E=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3E-E=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(2E=1-\frac{1}{3^8}\)

\(E=\frac{3^8-1}{2.3^8}\)

\(G=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\)

\(G=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{99}=\frac{1}{99}\)

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

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

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

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

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

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