a ) 

Theo bài ra: (a - 4) chia hết cho 5 => (a - 4) + 20 chia hết cho 5 => a + 16 chia hết cho 5

(a - 5) chia hết cho 7 => (a - 5) + 21 chia hết cho 7 => a + 16 chia hết cho 7

(a - 6) chia hết cho 11 => (a - 6) + 22 chia hết cho 11 => a + 16 chia hết cho 11 

=> a + 16 thuộc BC(5; 7; 11) 

Mà BCNN(5; 7; 11) = 385

=> a + 16 thuộc B(385) = {0; 385; 770; ...}

=> a thuộc {-16; 369; 754;...}

Vì a là số tự nhiên nhỏ nhất

=> a = 369 

b ) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}.\)

Ta có : 

\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

.....................

\(\frac{1}{2012^2}=\frac{1}{2012.2012}< \frac{1}{2011.2012}\)

Ta có :

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1-\frac{1}{2012}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}.< \frac{2011}{2012}\)

Mà \(\frac{2011}{2012}< 1\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\)

\(b)\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)​

\(< \)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{2010.2011}+\frac{1}{2011.2012}\)

\(< \)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(< \)\(1-\frac{1}{2012}\)\(=\frac{2011}{2012}< 1\)

Vậy Biểu thức    \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)\(< 1\)

\(a)\)

Theo bài ra: \(\left(a-4\right)⋮5\Rightarrow\left(a-4\right)+20⋮5\Rightarrow a+16⋮5\)

\(\left(a-5\right)⋮7\Rightarrow\left(a-5\right)+21⋮7\Rightarrow a+16⋮7\)

\(\left(a-6\right)⋮11\Rightarrow\left(a-6\right)+22⋮11\Rightarrow a+16⋮11\)

\(\Rightarrow\) \(a+16\in BC\left(5;7;11\right)\)

Mà \(BCNN\left(5;7;11\right)=385\)

\(\Rightarrow\) \(a+16\in B\left(385\right)=\left\{0;385;770;...\right\}\)

\(\Rightarrow\) \(a\in\left\{-16;369;754;...\right\}\)

Vì a là số tự nhiên nhỏ nhất \(\Rightarrow\) \(a=369\)

1a Để \(\frac{x+1}{2}\)=\(\frac{8}{x+1}\)

\(\Rightarrow\)x+1.(x+1)=2.8=16

\(\Rightarrow\)x+1(x+1)=4.4

suy ra x+1=4

x=4-1

x=3

a)(x+1)(x+1)=16

(x+1)^2=4^2

+)x+1=4

x=3

+)x+1=-4

x=-5

Bài 1 : 

1. a, 5\(^{2x-3}\)-2.5\(^2\)=5\(^2\).3

       5\(^{2x}\) : 5\(^3\) -2.25    = 25.3

       5\(^{2x}\):  5\(^3\) - 50      = 75

        5\(^{2x}\): 5\(^3\)            = 75+50

        5\(^{2x}\): 5\(^3\)            = 125

         5\(^{2x}\)                = 125.5\(^3\)

         5\(^{2x}\)                = 5\(^3\). 5\(^3\)

          5 \(^{2x}\)              = 5\(^{3+3}\)

          5 \(^{2x}\)               = 5\(^6\)  

Có 5=5 => 2x = 6

                  x = 6 : 2

                  x = 3

           Vậy x = 3.

b. / 2x -1 / = 5

=> 2x-1 = 5 hoặc 2x-1 = -5

* Với 2x - 1 = 5                                                                                      

thì     2x      = 5+1

        2x       = 6

          x       = 6:2

         x        = 3  

* Với 2x - 1 = - 5

thì     2x      = -5 + 1

         2x     = -4

           x      = -4 : 2

           x      = -2

câu b lên mạng có thể tìm thấy câu tương tự

Câu a ) 

S = 5 + 5² +..... + 5²⁰¹²

=> S \(⋮5\)

S = 5 + 5² +..... + 5²⁰¹²

S = ( 5 + 5³ ) + ( 5² + 5⁴ ) + ........ + ( 5²⁰¹⁰ + 5²⁰¹² )

S = 5 ( 1 + 5² ) + 5² ( 1 + 5² ) + ......... + 5²⁰¹⁰ ( 1 + 5² )

S = 5 x 26 + 5² x 26 + ................ + 5²⁰¹⁰ x 26

S = 26 ( 5 + 5² + .... + 5²⁰¹⁰ )

=> S\(⋮26\)

=>\(S⋮13\)( do 26 = 13 x 2 )

Do ( 5 , 13 ) = 1

=> \(S⋮5x13\)

=> \(S⋮65\)

1. 6300

54444

\(b.\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{4}{7}+\frac{3}{5}+\frac{1}{3}\)

\(\Leftrightarrow\frac{35+9}{105}< \frac{x}{210}< \frac{60+63+35}{105}\)

\(\Leftrightarrow\frac{44}{105}< \frac{x}{210}< \frac{158}{105}\)

\(\Leftrightarrow\frac{88}{210}< \frac{x}{210}< \frac{316}{210}\)

Suy ra \(x\in\left\{89;90;100;...;313;314;315\right\}\)

\(c.\left(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{19.21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(\Leftrightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+\frac{221}{231}=\frac{4}{3}\)

\(\Leftrightarrow\frac{1}{11}-\frac{1}{21}-x+\frac{221}{231}=\frac{4}{3}\)

\(\Leftrightarrow\frac{21-11-231x+221}{231}=\frac{308}{231}\)

\(\Leftrightarrow-231x=308-21+11-221\)

\(\Leftrightarrow-231x=77\)

\(\Leftrightarrow x=-\frac{77}{231}=-\frac{1}{3}\)

^^

Câu 4:

Ta có:

\(\frac{1}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\)

\(\frac{1}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\)

\(...\)

\(\frac{1}{98.99.100}=\frac{1}{98.99}-\frac{1}{99.100}\)

\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Rightarrow\frac{1}{k}=1\Rightarrow k=1:1=1\)