A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)

Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+..+\)\(\frac{1}{99.100}=\)\(1-\frac{1}{100}< 1\)

Mà A=1+B=>A=1+B<1+1=2

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)

B)

ta có : \(1=1\)

\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)

\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)

\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)

tất cả công lại \(\Rightarrow B< 6\)

a)A<1+1/1.2 +1/2.3 +1/3.4+...+1/99.100

A<1+1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

A<2-1/100<2

b)B=1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+...+1/16)+(1/17+1/18+...+1/32)+(1/33+1/34+...+1/63+1/64)-1/64

B<1+1/2+1/2+1/2+1/2+1/2+1/2-1/64

B<1+3-1/64

B<4-1/64<6

b, Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                \(\frac{1}{3^2}< \frac{1}{2.3}\)

                ..................

                 \(\frac{1}{100^2}< \frac{1}{99.100}\)

Nên C < \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\)

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

<=> C < \(1+1-\frac{1}{100}\)

<=> C < \(2-\frac{1}{100}=\frac{199}{100}\)

\(B=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+...+\left(\frac{1}{2^5}+...+\frac{1}{2^6-1}\right)\)

\(B< 1+\frac{1}{2}.2+\frac{1}{4}.4+...+\frac{1}{2^5}.32\)

\(B< 1+1+1+...+1\)( 6 số 1)

B<1.6=6

\(C=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)

\(C< 1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.10}\right)=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)\(=1+\left(1-\frac{1}{100}\right)< 1+1=2\)

Vậy C<2

C<1+1/1.2+1/2.3+...+1/99.100

=>C<1+1/1-1/100

<=>C<1+99/100

=>C<199/200

=>C<2

BẠN TỰ GIẢI CHI TIẾT HƠN NHA ^_^ 

MÌNH GIẢI HƠI TẮT CHÚT :))

b) A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

   3A=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

3A-A=\(1-\frac{1}{3^{99}}\)

   2A=\(1-\frac{1}{3^{99}}\)

vì 2A<1

=> A<\(\frac{1}{2}\)

Câu a) Mik chữa lại một chút 

Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\);.......; \(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

Suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

Suy ra: \(VT< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)

Vậy : \(VT+1< 1+1=2\)

Bài 1 :

\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)

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

\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)

\(\Rightarrow x\cdot\frac{24}{50}=1\)

\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)

                            #Louis

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

\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)

\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)

\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)

\(\frac{12}{25}x=1\)

Đến đây dễ rồi :)))

Bn tự tính típ nha

Bài 1 :

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

=> \(x\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)

=> \(x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)

=> \(x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)

=> \(x.\frac{12}{25}=1\)

=> \(x=\frac{25}{12}\)

Study well ! >_<

a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)

Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)

\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy...

b, Đặt A là tên của tổng trên

Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)

Đặt B là biêu thức trong ngoặc

Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow B< 2-\frac{1}{50}< 2\)

Thay B vào A ta được:

\(A< \frac{1}{2^2}.2=\frac{1}{2}\)

c, Đặt C = \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

\(C=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)

\(C=\left(1+1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)

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

Đặt D là biểu thức trong ngoặc

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)

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

\(\Rightarrow-D>-1\)

=>\(C=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1=48\)

Vậy C > 48

Y Ribi Nkok Ngok Lê Nguyễn Ngọc Nhi Lê Anh Duy Nguyễn Thị Diễm Quỳnh trần thị diệu linh kudo shinichi Nguyen Giang Thủy Tiên Nguyễn Việt Lâm