\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\); \(\frac{1}{4^2}< \frac{1}{3\cdot4}\); ....; \(\frac{1}{9^2}< \frac{1}{8\cdot9}\)

\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)

\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(\Rightarrow S< 1-\frac{1}{9}\)

\(\Rightarrow S< \frac{8}{9}\)    (1)

\(\frac{1}{2^2}>\frac{1}{2\cdot3};\frac{1}{3^2}>\frac{1}{3\cdot4};\frac{1}{4^2}>\frac{1}{4\cdot5};...;\frac{1}{9^2}>\frac{1}{9\cdot10}\)

\(\Rightarrow S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)

\(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Rightarrow S>\frac{1}{2}-\frac{1}{10}\)

\(\Rightarrow S>\frac{2}{5}\)   (2)

(1)(2) => 2/5 < S < 8/9

\(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}< \frac{1}{a^2}\)

\(\frac{1}{a}-1-\frac{1}{a}=-1< \frac{1}{a^2}\) Vì \(\frac{1}{a^2}>0;-1< 0\)

Khi đó thì ĐỀ SAI

A<1-1/2+1/2-1/3+...+1/8-1/9=1-1/9=8/9 A>1/2-1/3+1/3-1/4+...+1/9-1/10=1/2-1/10=2/5 =>2/5<A<8/9

bài này khó quá

A =\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{20^2}=\frac{1}{2^2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)

\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=\frac{1}{4}\left(1+1-\frac{1}{20}\right)=\frac{1}{4}\left(2-\frac{1}{20}\right)=\frac{1}{2}-\frac{1}{80}< \frac{1}{2}\left(\text{đpcm}\right)\)

a, với  n \(\in\) N, cm : 10ⁿ - 1 \(⋮\) 9

cách 1: Xét tổng các chữ số của hiệu 10ⁿ - 1 ta có 

1 + 0ⁿ - 1 = 0 \(⋮\) 9 \(\Rightarrow\) 10ⁿ - 1 \(⋮\) 9 (đpcm)

Cách 2 : Dùng đồng dư thức 

    10 \(\equiv\) 1 (mod 9) \(\Rightarrow\) 10ⁿ \(\equiv\)  1ⁿ (mod 9) \(\Rightarrow\) 10ⁿ - 1 \(\equiv\) 1ⁿ - 1 (mod 9) 

\(\Rightarrow\) 10ⁿ - 1 \(\equiv\) 0 (mod 9 ) \(\Rightarrow\) 10ⁿ - 1 \(⋮\) 9.

b, n \(\in\) N, cm : 10ⁿ + 8 \(⋮\) 9

Xét tổng các chữ số của tổng 10ⁿ + 8 ta có :

1 + 0ⁿ + 8 = 9 \(⋮\) 9 \(\Rightarrow\) 10ⁿ + 8 \(⋮\) 9 (đpcm)

mình cần gấp

chứng minh rằng : 5 ⁶ < 2 ¹⁴ < 5 ⁸

5⁶ = 5.5.5.5.5.5 = 15625

2 ¹⁴= 2.2.2.2.2.2.2.2.2.2.2.2.2.2=16384

5 ^{8 =}5.5.5.5.5.5.5.5=390625

mà 390625 > 16384 > 15625  = > 5 ⁶ < 2 ¹⁴ < 5 ⁸

#HỌCTỐT
&YOUTUBER&
 

1/a,

-Ta có: 

$B<1\Leftrightarrow B<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10(10^{2004}+1)}{10(10^{2005}+1)}=\frac{10^{2004}+1}{10^{2005}+1}=A$

-Vậy: B<A

b,$A=1+(\frac{1}{2})^2+...+(\frac{1}{100})^2$

$\Leftrightarrow A=1+\frac{1}{2^2}+...+\frac{1}{100^2}$

$\Leftrightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$

$\Leftrightarrow A<1+\frac{1}{1}-\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}$

$\Leftrightarrow A<1+1-\frac{1}{100}\Leftrightarrow A<2-\frac{1}{100}\Leftrightarrow A<2(đpcm)$
2,
a.
-Ta có:$\Rightarrow \frac{3x+7}{x-1}=\frac{3(x-1)+16}{x-1}=\frac{3(x-1)}{x-1}+\frac{16}{x-1}=3+\frac{16}{x-1}
-Để: 3x+7/x-1 nguyên
-Thì: $\frac{16}{x-1}$ nguyên
$\Rightarrow 16\vdots x-1\Leftrightarrow x-1\in Ư(16)\Leftrightarrow ....$
b, -Ta có:
$\frac{n-2}{n+5}=\frac{n+5-7}{n+5}=1-\frac{7}{n+5}$
-Để: n-2/n+5 nguyên
-Thì: \frac{7}{n+5} nguyên
$\Leftrightarrow 7\vdots n+5\Leftrightarrow n+5\in Ư(7)\Leftrightarrow ...$

ta có 

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

hay 

\(4A=A+4^{100}-1\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}=\frac{B}{3}\)

vậy ta có điều phải chứng minh

a) 10^n-1=100...0(n chữ số 0)-1=999...9(n chữ số 9) chia hết cho 9

b)10^n+8=100...0(n chữ số 0)+8=100....08(n-1 chữ số 0) chia hết cho 9