\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{43.46}\\ S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}\\ S=1-\dfrac{1}{46}< 1\)

Vậy S < 1 (đpcm)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{43}-\frac{1}{46}..\)

\(S=1-\frac{1}{46}< 1\)

VẬY S<1

\(S=\frac{3}{1.4} +\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

=> S<1 (ĐCCM)

Bạn có cần giải thích tại sao tách được các phân số như thế không?

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{10}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}\)

Vì \(1-\frac{1}{46}\) < 1

=> \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1

Như trên

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

Có \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\)

nhan xet:3/1.4=1/1-1/4

3/4.7=1/4-1/7

3/7.10=1/7-1/10

.....................

3/40.43=1/40-1/43

3/43.46=1/43-1/46

S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S=1/1-1/46

S=46/46-1/46

S=45/46<1

vay s<1

Giải:

a) Ta có:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

1/5²=1/5.5 < 1/4.5

1/6²=1/6.6 < 1/5.6

1/7²=1/7.7 < 1/6.7

1/8²=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

   B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

   B<1/1-1/8

   B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

   S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

   S=1/1-1/46

   S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!

a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)

\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)

Vậy B < 1 (đpcm)

S=1 - 1/46 < 1 

Có \(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{43.46}\)

Sẽ có: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{43.46}\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+........+\frac{1}{43}-\frac{1}{46}\)

\(S=\frac{1}{1}-\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+....+\left(-\frac{1}{43}+\frac{1}{43}\right)-\frac{1}{46}\)

\(S=\frac{1}{1}-0+0+0+0+......+0+0-\frac{1}{46}\)

\(S=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)

Vì có:  \(\frac{45}{46}<1\) nên \(S<1\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

\(S=\frac{45}{46}\)

=> \(\frac{45}{46}<\frac{46}{46}=1\)

=> S<1

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

ta có 

\(\frac{3}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{1}{4}-\frac{1}{7}\)

.....

\(\frac{3}{43.46}=\frac{1}{43}-\frac{1}{46}\)

( mình nghĩ cậu chưa đc làm dạng như này nên ghi ra , lần sau có gặp mà biết cách làm  r thì bỏ bước trên đi cx đc nha)

\(=>S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=>S=1-\frac{1}{46}< 1\)(dpcm

S= 1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+|1/43-1/46

S= 1-1/46

S= 45/46<1

vậy S<1

duyệt đi

S=  1- 1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S= 1+ (1/4-1/4)+(1/7-1/7)+...+(1/43-1/43)-1/46

S= 1-1/46= 45/46<1

Suy ra S<1