S = 22 + 42 + 62 + ... + 202

   = (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2

   = 22.12 + 22.22 + 22.32 + ... + 22.102

   = 22 (12 + 22 + ... + 102 )

   = 4 . 385

   = 1540

= (1x2)^2 (2x2)^2 (3x2)^2 (4x2)^2 ..... (9x2)^2 (10x2)^2 
= 1^2 x 2^2 2^2 x 2^2 3^2 x 2^2 4^2 x 2^2 ..... 9^2 x 2^2 10^2 x 2^2 
= (1^2 2^2 3^2 4^2 ..... 9^2 10^2) x 2^2 
= 385 x 2^2 = 385 x 4 = 1540

giúp mình với

  4²:4.3³-2²+7
⇒ 4.27-4+7
⇒ 108-4+7
⇒ 111

    4\(^2\) : 4.3\(^3\) - 2\(^2\) + 7
= 16 : 4 . 27 - 4 + 7
= 4 . 27 - 4 + 7
= 108 - 4 + 7
= 104 + 7
= 111
-Tick cho mình nha mình cảm ơn ạ. <3

A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)

Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)

nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)

\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)

A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)

A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)

\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\) 

Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A>  \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)

Tương tự ta có :

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)

A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)

A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)

Ta thấy : 

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

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

......

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow S< 1-\frac{1}{100}\)

Mà \(1-\frac{1}{100}< 1\)nên \(S< 1\)

Ủng hộ mk nha !!! *_*

\(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(=\frac{99}{100}< 1\left(đpcm\right)\)

1/2^2+1/3^2+1/4^2+....+1/2005^2

ta có vì:1/2^2<1/2; 1/3^2 <1/2.....;1/2005^2<1/2

suy ra 1/2^2+1/3^2+1/4^2+....+1/2005^2<1/2

86:[2.(2.x-1)²-7]+4² = 2.3²

86:[2.(2.x-1)²-7]+16 = 2.9

86:[2.(2.x-1)²-7]+16 = 18

86:[2.(2.x-1)²-7]       = 18-16

86:[2.(2.x-1)²-7]       = 2

      2.(2.x-1)²-7        = 86:2

      2.(2.x-1)²-7        = 43

      2.(2.x-1)²           = 43+7

      2.(2.x-1)²           = 50

         (2.x-1)²           = 50:2

         (2.x-1)²           = 25

         (2.x-1)²           =5²

\(\Rightarrow\)2.x-1               = 5

        2.x                   = 5+1

        2.x                   = 6

           x                   = 6:2

          x                    = 3