câu này hình như có trong toán mạng

1/201>1/300,1/202>1/300.................1/300=1/300 =>S>1/300.100=1/3                                                 1/201<1/200, 1/202<1/300.................1/300<1/200=>S<1/200.100=1/2

Ta chia A ra lam 2 khoang la tu B=1/201+1/202+...+1/250 (50 so hang)va tu C=1/251+1/252+...+1/300 (50 so hang)

ta thay 1/201=1/201, 1/202<1/201,....1/250<1/201

cong  ca 2 ve cua 50 bat dang thuc cung chieu ta duoc B<50.1/201=> B<50/201

ta cung thay 1/251=1/251,1/252<1/251,....1/300<1/251

cong 2 ve cua 50 bat dang thuc cung chieu ta lai duoc C<50.1/251=> C<50/251

Ta thay A= B+C

suy ra A< 50/201+50/251=> A<0,448

ma tha thay 0,448<9/20

SUY RA A<9/20 (dpcm)

Thoi mik ko lay tien cua ban dau

Co gang len! Chuc ban hoc tot!

Really ? 

Có chắc là có 50K thật không ?? 

a: Ta có: \(\frac{1}{201}>\frac{1}{400};\frac{1}{202}>\frac{1}{400};\ldots;\frac{1}{400}=\frac{1}{400}\)

Do đó: \(\frac{1}{201}+\frac{1}{202}+\cdots+\frac{1}{400}>\frac{1}{400}+\frac{1}{400}+\cdots+\frac{1}{400}\)

=>\(\frac{1}{201}+\frac{1}{202}+\cdots+\frac{1}{400}>\frac{200}{400}=\frac12\)

b: Ta có: \(\frac{1}{201}<\frac{1}{200};\frac{1}{202}<\frac{1}{200};\ldots;\frac{1}{400}<\frac{1}{200}\)

Do đó: \(\frac{1}{201}+\frac{1}{202}+\cdots+\frac{1}{400}<\frac{1}{200}+\frac{1}{200}+\cdots+\frac{1}{200}\)

=>\(\frac{1}{201}+\frac{1}{202}+\cdots+\frac{1}{400}<\frac{200}{200}=1\)