M=1/10 + 1/15 + 1/21 +....+  1/120

M=2/20 +2/30+2/42+....+2/240

M=2/4.5 + 2/5.6 + 2/6.7 +.....+ 2/15.16

M=2.(1/4.5 +......+ 1/15.16)

M=2.(1/4 -1/5 +1/5 - 1/6 +.....+ 1/15 - 1/16)

M=2.(1/4 - 1/16)

M=2.(4/16 - 1/16)

M=2. 3/16

M=6/16=3/8 

Có 1/3 = 8/24 < 9/24 = 3/8 =>1/3<M

Có 1/2 = 4/8>3/8 =>1/2 >M

=> 1/3 < M < 1/2

cảm ơn bạn rất nhiều

k cho mik

a)     5/30+15/90+25/150+35/210+45/270

       =1/6+1/6+1/6+1/6+1/6

       =1/6 x 5

       =5/6

b)     1/2+1/6+1/12+1/20+....+1/56

        =1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8

        =1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8

        =1/1-1/8

         =7/8

c)     mình chịu

thank you bn nhìu nha

chịu,khó quá,nhất là câu c ý

#)Giải :

\(\frac{1}{15}+\frac{4}{30}+\frac{9}{45}+\frac{16}{60}+...+\frac{81}{135}=\frac{1}{15}+\frac{2}{15}+\frac{3}{15}+...+\frac{9}{15}=\frac{45}{15}=3\)

Dễ ẹc ak :v rút gọn là ra

=(\(\frac{1}{15}\)+\(\frac{4}{30}\)+\(\frac{16}{60}\)+\(\frac{64}{120}\))+(\(\frac{9}{45}\)+\(\frac{36}{90}\))+(\(\frac{25}{75}\)+\(\frac{81}{135}\))

=(\(\frac{8}{120}\)+\(\frac{16}{120}\)+\(\frac{32}{120}\)+\(\frac{64}{120}\))+(\(\frac{18}{90}\)+\(\frac{36}{90}\))+\(\frac{14}{15}\).

=1+\(\frac{3}{5}\)+\(\frac{14}{15}\).

=\(\frac{8}{5}\)+\(\frac{14}{15}\).

=\(\frac{15}{38}\)

Đặt A = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2015!}\)

A < \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2014.2015}\)

A < \(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2014}-\frac{1}{2015}\)

A < \(2-\frac{1}{2015}\)< 2 < \(2\left(\frac{135^2+136}{136^2-135}\right)\)

=> A < \(2\left(\frac{135^2+136}{136^2-135}\right)\)(Đpcm)

b, \(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

     \(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)

   \(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)

\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)

==> x+200=0

<=>x=-200

Vậy nghiệm của phương trình là x=-200

c,  \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

      \(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

mà  \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)

==>200-x=0

<=>x=200

vậy nghiệm của pt là x=200

a, \(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)

\(2\left(\frac{11x+4}{12}\right)-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-\frac{24}{12}+\frac{2x}{12}=0\)

\(\frac{44x+8-24+2x}{12}=\frac{46x-16}{12}=0\)

\(\Leftrightarrow46x-16=0\)

\(\Leftrightarrow46x=16\Rightarrow x=\frac{8}{23}\)

Vậy nghiệm của pt là x=8/23

k mk

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x = -100

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)

\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)

Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)

\(\Rightarrow200-x=0\Rightarrow x=200\)

Vậy x = 200