ta có : 2/3'2 < 2/2.3 ; 2/4'2<2/3.4 ... ;2/100'2<2/99.100

nen 2/3'2 +2/4'2+...+2/100'2<2/2.3+2/3.4+...+2/99.100 (1)

ta có  2/2.3+2/3.49+...+2/99.100

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

=2.(1/2-1/100)

=2.(50/100-1/100)

=2.49/100

ma 1>49/100

nen 1>1/2-1/3+...+1/99-1/100 (2)

tu(1) va (2) suy ra 2/3'2+...+2/100'2 >1

\(A=1^2+3^2+5^2+...+99^2\)

=>\(A=\left(1^2+2^2+...+99^2+100^2\right)-\left(2^2+4^2+...+100^2\right)\)

\(=\left(1^2+2^2+...+100^2\right)-4\left(1^2+2^2+...+50^2\right)\)

\(=\dfrac{100\cdot\left(100+1\right)\left(100\cdot2+1\right)}{6}-4\cdot\dfrac{50\cdot\left(50+1\right)\left(50\cdot2+1\right)}{6}\)

\(=166650\)

a) \(\left(4\frac{1}{2}-2x\right)\cdot3\frac{2}{3}=\frac{11}{5}\)

\(\left(\frac{9}{2}-2x\right)=\frac{11}{5}\cdot\frac{3}{11}\)

\(2x=\frac{45-6}{10}\)

\(2x=\frac{39}{10}\)

\(x=\frac{39}{10\cdot2}=\frac{39}{20}\)

b) \(\frac{3}{4}\cdot x+\frac{4}{7}\cdot x=-\frac{15}{8}\)

\(x\cdot\left(\frac{21+16}{28}\right)=-\frac{15}{8}\)

\(x=-\frac{15}{8}\cdot\frac{28}{37}\)

\(x=-\frac{105}{74}\)

i

help me

A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{812}\) + \(\dfrac{1}{870}\)

A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{28\times29}\)+ \(\dfrac{1}{29\times30}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)  - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+\(\dfrac{1}{28}\)-\(\dfrac{1}{29}\)+ \(\dfrac{1}{29}\) - \(\dfrac{1}{30}\)

A = 1 - \(\dfrac{1}{30}\)

A = \(\dfrac{29}{30}\)

29/30 nha

Chứng minh :

           \(S=\frac{1}{5}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{25}+\frac{1}{101}+\frac{1}{102}+...+\frac{1}{105}< \frac{1}{2}\) 

Nhóm các số hạng: 

           \(S=\frac{1}{5}+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{25}\right)+\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{105}\right)< \frac{1}{5}+\frac{5}{21}+\frac{5}{101}< \frac{1}{5}+\frac{5}{20}+\frac{5}{100}=\frac{1}{2}.\)

1)

A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

   = \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)

   = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)

   = \(\frac{1}{5}-\frac{1}{12}\)

   = \(\frac{7}{60}\)

B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

   = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

   = \(\frac{3.4.5.....100}{2.3.4....99}\)

   = \(\frac{100}{2}=50\)

C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)

   = \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)

   = \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)

   = \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)

   = \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

~ Hok tốt ~

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

....

`1/3+ -1/4+1/5+ -1/6+1/7+1/6+1/(-5)+1/4+ -1/3`

`=(1/3-1/3)+(1/4-1/4)+(1/5-1/5)+(1/6-1/6)+1/7`

`=0+0+0+0+1/7`

`=1/7`

gấp ạ