đặt A=1/1.2+1/2.3+1/3.4+..........1/49.50

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}<1\)

vậy A<1

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

1 - 1/50 < 1

1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/49.50

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/49 - 1/50

1 - 1/50 < 1

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

=\(1-\frac{1}{50}\)

Vì \(1-\frac{1}{50}< 1\)nên A < 1

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

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

=\(\frac{1}{2}-\frac{1}{100}\)

Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(\Rightarrow A< 1\)

\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=\frac{1}{2}-\frac{1}{100}\)

\(\Rightarrow B< \frac{1}{2}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}\)

\(A=\frac{49}{50}< 1\)

\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)

\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)

M=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50<1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{49.50}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)

\(M=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+........+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)

\(M=\frac{1}{1}-0+0+0+0+0+......+0+0-\frac{1}{50}\)

\(M=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)

Vì \(\frac{49}{50}<1\) nên  \(S<1\)

\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

     \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

     \(=1-\frac{1}{50}<1\)

\(\Rightarrow M<1\) 

Vậy \(M<1\)

Chúc bạn học tốt!!!!!!!

M=1/1.2+1/2.3+1/3.4+...+1/49.50

M=1-1/2+1/2-1/3+...+1/49-1/50

M=1-1/50<1

Vậy M<1

M=1/1.2+1/2.3+...+1/49.50

M=1/1-1/2+1/2-1/3+.....+1/49-1/50

M=1-1/50<1

=>M<1

\(M=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)

\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(M=1-\frac{1}{50}<1\)

\(=>M<1\)

M = 1/1.2 + 1/2.3 + ... + 1/49.50

M = 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50

M = 1 - 1/50

M = 49/50 

Vì 49/50 < 1

=> M < 1

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}=\frac{49}{50}\)

\(B=1.2+2.3+3.4+...+49.50\)

\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(=49.50.51\)

\(B=\frac{49.50.51}{3}=49.50.17\)

\(50^2.A-\frac{B}{17}=49.50-49.50=0\)

1/1.2   +  1/2.3   +  1/3.4  + ... +   1/49.50

=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50

=49/50 

ta có :9/10=45/50

=>49/50>45/50

=>1/1.2   +  1/2.3   +  1/3.4  + ... +   1/49.50  > 9/10

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1 - 1/50

= 49/50

=> 49/50 > 9/10

=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 > 9/10

Chúc bn có kết quả hc kì II thật tốt nha !!!!!!!!!!!! ^_^

mình làm luôn nha :

    Ta có : 1/1.2+1/2.3+1/3.4+.....+1/49.50 =1-1/2+1/2-1/3=1/3-1/4+....+1/49-1/50

                                                             =1-1/50=49/50

Vì 49/50>45/50 = 9/10 nên 45/50>9/10

                                     Vậy 1/1.2+1/2.3+1/3.4+...+1/49.50 > 9/10

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

thấy công thức trên vào biểu thức, khử liên tiếp, ta con

1-1/50  <1

Ta cộng vào biểu thức trên( đặt là A) 1 dãy là:1/2*3+1/4*5+1/6*7+...+1/47*48.(đặt là B).

=>A+B>A.

Ta có:A+B= 1/1*2+1/2*3+1/3*4+1/4*5+...+1/49*50.

=>A+B=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50.

=>A+B=1-1/50.

=>A+B<.

Mà A+B>A=>A<1.

Vậy A<1.

tk nha đúng 1000000% .

-chúc các bạn tk mk học giỏi nha-