nó dài lắm mik lười giải

Lời giải:
$0,65=\frac{65}{100}=\frac{65:5}{100:5}=\frac{13}{20}$

Đáp án D.

Đề thiếu số đo góc A rồi em!

Gọi tổng cần tính là A.

Ta có:

\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\\ A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\\ 2A=2\cdot\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\\ 2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\\ A=2-\dfrac{1}{2^{10}}\)

Lời giải

$M.\frac{1}{2}=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{210}+\frac{1}{240}$

$=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{14.15}+\frac{1}{15.16}$
$=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{15-14}{14.15}+\frac{16-15}{15.16}$

$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}$

$=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$

$\Rightarrow M=\frac{3}{8}$

Vì $\frac{1}{3}< \frac{3}{8}< \frac{1}{2}$ nên $\frac{1}{3}< M< \frac{1}{2}$

 M=1/10 + 1/15 + 1/21 + 1/28 +....+ 1/105 + 1/120                                 M=1/2*5 + 1/5*3 +1/3*7 + 1/7*4 +....+ 1/7*15 + 1/15*8                     M=2(1/2*2*5+1/2*5*3+1/2*3*7+1/2*7*4+....+1/2*7*15+1/2*15*8)   M=2(1/4*5+1/5*6+1/6*7+1/7*8+....+1/14*15+1/15*16)                   M=2(1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+....+1/14-1/15+1/15-1/16)       M=2(1/4-1/16)                                                                                     M=2(4/16-1/16)                                                                                   M=2*3/16                                                                                               M=3/8

Bài 1:

\(E=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{9894}\\ =\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{97.102}\)

\(\Rightarrow 5E=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+...+\frac{102-97}{97.102}\\ =1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{102}\\ =1-\frac{1}{102}=\frac{101}{102}\\ \Rightarrow E=\frac{101}{102}:5=\frac{101}{510}\)

Bài 2:

Gọi số học sinh cả lớp là $a$

Số học sinh vắng mặt ban đầu: $\frac{a}{1+6}.1=\frac{a}{7}$ (hs) 

Sau khi 2 học sinh ra khỏi lớp thì lớp còn $a-2$ hs

Số hs vắng mặt lúc sau: $\frac{a-2}{1+4}.1=\frac{a-2}{5}$ (hs)

Có: 

$\frac{a-2}{5}=\frac{a}{7}+2=\frac{a+14}{7}$

$\Rightarrow 7(a-2)=5(a+14)$

$\Rightarrow 7a-14=5a+70$

$\Rightarrow 2a=84$
$\Rightarrow a=42$ (hs)

Đặt \(A=1+2+2^2+...+2^{2009}\)

=>\(2A=2+2^2+2^3+...+2^{2010}\)

=>\(2A-A=2+2^2+...+2^{2010}-1-2-...-2^{2009}\)

=>\(A=2^{2010}-1\)

\(\dfrac{A}{1-2^{2010}}=\dfrac{2^{2010}-1}{1-2^{2010}}=-1\)

Câu hỏi đâu bạn ? 

$A=1.21+3.41+...+49.501$ hiển nhiên $>1$ rồi mà bạn. Bạn xem lại đề.

A = ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{5.6}\)} + ... + ^{\(\dfrac{1}{49.50}\)} 

A <  ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{5.6}\)} + ... + ^{\(\dfrac{1}{49.50}\)} +^{\(\dfrac{1}{2.3}+\dfrac{1}{4.5}+\dfrac{1}{6.7}\)+...+\(\dfrac{1}{48.49}\)}

A < ^{\(\dfrac{1}{1.2}\)} + ^{\(\dfrac{1}{2.3}\)} + ^{\(\dfrac{1}{3.4}\)} + ^{\(\dfrac{1}{4.5}\)}+^{\(\dfrac{1}{5.6}\)} +^{\(\dfrac{1}{6.7}\)}+.. +^{\(\dfrac{1}{47.48}\)}+ ^{\(\dfrac{1}{48.49}\)}+ ^{\(\dfrac{1}{49.50}\)} 

A < ^{\(\dfrac{1}{1}\)}-^{\(\dfrac{1}{2}\)} + ^{\(\dfrac{1}{2}\)} - ^{\(\dfrac{1}{3}\)} + ^{\(\dfrac{1}{3}\)} - ^{\(\dfrac{1}{4}\)} + ... + ^{\(\dfrac{1}{49}\)} - ^{\(\dfrac{1}{50}\)}

A < ^{\(\dfrac{1}{1}\)} - ^{\(\dfrac{1}{50}\)} < 1 (đpcm)

Lời giải:

$\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{60^2}$

$=\frac{1}{9}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{60^2}

$< \frac{1}{9}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{59.60}$

$= \frac{1}{9}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{60-59}{59.60}$

$=\frac{1}{9}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}$

$=\frac{1}{9}+\frac{1}{3}-\frac{1}{60}=\frac{4}{9}-\frac{1}{60}< \frac{4}{9}$

Ta có đpcm.