\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)

\(suyra:A<1\)

bạn ơi tách ra thừa số chung rồi làm như bình thường nha 

1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)

     =4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)

     = 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)

    =4-\(\left(1-\frac{1}{9}\right)\)

     = 4-\(\frac{8}{9}\)

      = \(\frac{7}{9}\)

Câu 2 tương tự như câu 1 

A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{154}\right)\)

A= 4                    -\(\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{11.14}\right)\)

Bạn tự làm tiếp 

\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{184}+\frac{1}{238}+\frac{1}{340}=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}>\frac{2}{20}=\frac{1}{10}=0,1\)

vậy S>0,1

S = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

S = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

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

S = \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}\)

S = \(\frac{3}{20}\)

S = 0,15 > 0,1

S=\(\frac{3}{20}\)>0,1

li ke cho mình nha

Ta có : 

\(N=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(N=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(3N=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(3N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(3N=\frac{1}{2}-\frac{1}{20}\)

\(3N=\frac{9}{20}\)

\(N=\frac{9}{20}:3\)

\(N=\frac{3}{20}\)

Vậy \(N=\frac{3}{20}\)

Chúc bạn học tốt ~ 

\(N=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{238}+\frac{1}{340}\)

\(N=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{14.17}+\frac{1}{17.20}\)

\(N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(N=\frac{1}{2}-\frac{1}{20}\)

\(N=\frac{10}{20}-\frac{1}{20}\)

\(N=\frac{9}{20}\)

\(N=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(N=\frac{1}{2x5}+\frac{1}{5x8}+\frac{1}{8x11}+\frac{1}{11x14}+\frac{1}{14x17}+\frac{1}{17x20}\)

\(N=\frac{1}{3}x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(N=\frac{1}{3}x\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(N=\frac{1}{3}x\frac{9}{20}\)

\(N=\frac{3}{20}\)

a, 1/2x5 - 1/5x8 - 1/8x11 - 1/11x14

= 1/2 x 1/5 - 1/5 x 1/8 - 1/8 x 1/11 - 1/11 x 1/14

= 1/2 - 1/14 = 3/7

b và c mik nghĩ là lỗi đề (vì mik đã được học r). Theo mik nghĩ đề phải như thế này:

b) 1/2 + 1/6 + 1/12 + ... + 1/56 + 1/72

c) 9/10 + 39/40 + 87/88 + 153/154

@Đỗ Khánh Minh 

Cảm ơn bạn nhiều

Bài 2:

a. $=62-81-12+59-9=(62-12)+(59-9)-81$

$=50+50-81=100-81=19$

b. $=39+13-26-62-39=(39-39)+13-(26+62)$

$=0+13-88=-(88-13)=-75$

c. $=(32-42)+(36-34)+(40-38)=10+2+2=14$
d. $=92-55+8-45=(92+8)-(55+45)=100-100=0$

Bài 1:

a. $=(387-87)-224=300-224=76$

b. $=-(75+35)+379=-110+379=379-110=269$

c. $=(11+15)-(13+17)=25-30=-5$

d. $=(31-21)-(27-24)=10-3=7$