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)

2x(3y-2)+(3y-2) = (2x+1)(3y-2) = -55.Lập bảng :

Vậy (x;y) = (-28;1);(-1;19);(2;-3);(5;-1)

làm giúp mình câu b) nhé ! cảm ơn bạn nhiều !!!

   D = 4⁰ + 4¹ + 4² + 4³ + 4⁴ + ... + 4²⁰⁰

4.D = 4   + 4² + 4³ + 4⁴ + 4⁵ +... + 4²⁰¹

4D - D = (4 + 4² + 4³ + 4⁴ + 4⁵ + ... + 4²⁰¹) - (4⁰ + 4¹ + 4² +...+4²⁰⁰)

 3D      = 4 + 4² + 4⁴ + 4⁴ + 4⁵ + ... + 4²⁰¹ - 4⁰ - 4¹ - 4² - ... - 4²⁰⁰

3D      =  (4 - 4¹) + (4² - 4²) + .... + (4²⁰⁰ - 4²⁰⁰) + 4²⁰¹ - 4⁰

  3D    = 4²⁰¹ - 4⁰

   3D + 1 = 4²⁰¹ - 1 + 1

   3D + 1  = 4²⁰¹

Theo bài ra ta có: 4²⁰¹ = 4ⁿ⁺¹

                              n + 1  = 201

                              n =  201 - 1

                              n = 200

\(D=4^0+4^1+4^2+4^3+4^4+...+4^{200}\\4D=4\cdot(4^0+4^1+4^2+4^3+4^4+...+4^{200})\\4D=4^1+4^2+4^3+4^4+4^5+...+4^{201}\\4D-D=(4^1+4^2+4^3+4^4+4^5+...+4^{201})-(4^0+4^1+4^2+4^3+4^4+...+4^{200})\\3D=4^{101}-4^0\\3D=4^{101}-1\\\Rightarrow 3D+1=4^{101}\)

Mặt khác: \(3D+1=4^{n+1}\)

\(\Rightarrow 4^{n+1}=4^{101}\\\Rightarrow n+1=101\\\Rightarrow n=101-1=100(tmdk)\)

D=1+4+4²+4³+4⁴+...+4²⁰⁰

4D=4+4²+4³+4⁴+4⁵+...+4²⁰¹

4D-D=4²⁰¹-1

3D=4²⁰¹-1

3D+1=4²⁰¹-1+1

3D+1=4²⁰¹

Ý tưởng: 4²⁰¹=4ⁿ⁺¹

201=n+1

n+1=201

n=201-1

n=200

(đpcm)

A=1/4(1/1+1/2^2+...+1/50^2)

=>A=1/4+1/4*(1/2^2+...+1/50^2)

=>A<1/4+1/4*(1-1/2+1/2-1/3+...+1/49-1/50)

=>A<1/4+1/4*49/50=99/200<1/2

1)Tính:

a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)

b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)

c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)

a

\(4^2.2=16.2=32\)

b\(36^2:6^2=36.36:6.6=36.36:36=36\)

c

3)

a)\(x:\left(-\frac{1}{3}\right)^5=\left(-\frac{1}{3}\right)^3\)

                      \(x=\left(-\frac{1}{3}\right)^3\cdot\left(-\frac{1}{3}\right)^5\)           

                      \(x=\left(-\frac{1}{3}\right)^8\)

b)\(\left(\frac{2}{3}\right)^{16}\cdot x=\left(\frac{2}{3}\right)^{18}\)

                      \(x=\left(\frac{2}{3}\right)^{18}:\left(\frac{2}{3}\right)^{16}\)

                      \(x=\left(\frac{2}{3}\right)^2\)

                      \(x=\frac{2^2}{3^2}\)

                      \(x=\frac{4}{9}\)

a)6²:4.3+2.5²

=36:4.3+2.25

=27+50

=77

Mà: 77=7.11

b) 6.4²-18.3²

=6.16-18.9

=96-162

= -96

rút 2 làm thừa số chung