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

\(>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\)

\(=\frac{100}{200}=\frac{1}{2}\)

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

\(< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)

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

Suy ra dpcm. 

ta có:\(B=\frac{10n-3}{4n-10}=\frac{5.\left(2n-5\right)+22}{2.\left(2n-5\right)}=\frac{5}{2}+\frac{22}{2.\left(2n-5\right)}=\frac{5}{2}+\frac{11}{2n-5}\)

\(Bmax\Leftrightarrow\frac{5}{2}+\frac{11}{2n-5}max\Leftrightarrow\frac{11}{2n-5}max\Leftrightarrow2n-5=1\)

\(\Leftrightarrow2n=6\Leftrightarrow n=3\)

\(B=\frac{5}{2}+11=\frac{27}{2}\)

VẬY \(n=3\) THÌ \(maxB=\frac{27}{2}\)

Bằng 1141451 :)

2487878 - 1346427 = 1141451 nhé

TL

=37627200000000000000

nha bn

HT

-15/3<x<21/7

-3<x<3

x=2 (số nguyên tố lớn hơn 1)

HT

TL

X=3

nha bn

HT

1/4 - 111/444 = 0

1/4-111/444

=1/4-1/4

=0

_HT_

Gọi biểu thức này là \(A\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{302^2}\)

Với mọi số tự nhiên ta luôn có : \(\left(n-1\right)^2< n^2< \left(n+1\right)^2\)

\(\Rightarrow\frac{1}{\left(n+1\right)^2}< \frac{1}{n^2}< \frac{1}{\left(n-1\right)^2}\)

Áp dụng ta có :

\(\Rightarrow A< \frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+......+\frac{1}{30^2}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{29.30}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{29}-\frac{1}{30}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{30}< \frac{1}{2}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)

TA CÓ:

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{30^2}< \frac{1}{29.30}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)

ta lại có:\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{30-29}{29.30}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{29}-\frac{1}{30}\)

\(=1-\frac{1}{30}< 1\)

mà \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{29.30}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}< 1-\frac{1}{30}< 1\left(đpcm\right)\)

Ta có : \(-\frac{92}{111}< -\frac{4}{5}\)

\(-\frac{4}{5}< \frac{35}{-44}\)

\(\Rightarrow-\frac{92}{111}< \frac{35}{-44}\)

_HT_

\(\frac{-92}{111}\)\(=\frac{-4}{5}\)

\(\frac{-4}{5}\)\(< \frac{35}{-44}\)

\(\Rightarrow< \)