Ta thấy tổng trên có 50 số hạng .

Ta có:

1/101>1/150

1/102>1/150

...

1/149>1/150

1/150=1/150

=>1/101+1/102+...+1/149+1/150>1/150+1/150+...+1/150

                                                 ---50 số hạng 1/150-------

=>1/101+1/102+...+1/149+1/150>1/150.50

=>1/101+1/102+...+1/149+1/150>50/150

=>1/101+1/102+...+1/149+1/150>1/3

em lạy chị Nguyễn Trà My cho em **** đi mà

ta có: 1/101>1/150

          1/102>1/150

           ...................

           1/150=1/150

=> 1/101+1/102+1/103+...+1/150>1/150+1/150+1/150+...+1/150

                                                         _tổng 1/150 có 50 số hạng_

=>1/101+1/102+1/103+...+1/150>1/150.50

=>1/101+1/102+1/103+...+1/150>50/150

=>1/101+1/102+1/103+...+1/150>1/3

\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\)(50 phân số)

=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\)(50 phân số)

=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50\)

=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{3}\)(Đpcm)

Ta có :

\(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}=VP\left(đpcm\right)\)

Xét :

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)

Thêm \(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\)vào mỗi vế ta có

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)

\(\RightarrowĐPCM\)

Ta có:

1/101 + 1/102 + ... + 1/149 + 1/150 > 1/150 + 1/150 + ... + 1/150 + 1/150  

                                                                                      50 phân số 1/150

                                                                 > 50.1/150

                                                                > 1/3

=>  đpcm

ta có

\(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};....;\frac{1}{150}=\frac{1}{150}\)

=>tổng>\(\frac{1}{150}.50=\frac{1}{3}\)

=>đpcm

RÚT GỌN:

\(\frac{3.13-13.18}{15.40-80}=\frac{13.\left(3-18\right)}{15.40-40.2}=\frac{13.\left(-15\right)}{40\left(15-2\right)}=\frac{13.\left(-15\right)}{40.13}=-\frac{15}{40}=-\frac{3}{8}\)

CHỨNG MINH:

Ta thấy \(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};...;\frac{1}{149}>\frac{1}{150}\)

=>\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}.\left(150-101+1\right)=\frac{1}{150}.50=\frac{50}{150}=\frac{1}{3}\)(đpcm)

TÍNH HỢP LÝ:

B=\(\frac{5}{13}+\frac{-5}{7}-\frac{20}{41}+\frac{8}{13}+\frac{-21}{41}=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-21}{41}-\frac{20}{41}\right)+\frac{-5}{7}=1+\left(-1\right)+\frac{-5}{7}=0+\frac{-5}{7}=\frac{-5}{7}\)

Tách A thành 2 nhóm A₁ , A₂

A₁ = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\)

A₂ = \(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{200}>\frac{1}{200}.50=\frac{1}{4}\)

\(\Rightarrow\)A = A₁ + A₂ > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)