\(B=\dfrac{2014}{1}+\dfrac{2013}{2}+...+\dfrac{1}{2014}\)

\(=\left(\dfrac{2013}{2}+1\right)+\left(\dfrac{2012}{3}+1\right)+...+\left(\dfrac{1}{2014}+1\right)+1\)

\(=\dfrac{2015}{2}+\dfrac{2015}{3}+...+\dfrac{2015}{2014}+\dfrac{2015}{2015}\)

\(=2015\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)

=2015A

=>\(\dfrac{A}{B}=\dfrac{1}{2015}\)

b.-9/7

Chọn B

                     Giải

Gọi số trứng là \(x\) ( quả trứng); \(x\in\) N*

Số trứng bán ngày thứ nhất là: \(x\) x \(\dfrac{1}{4}\) (quả trứng)

Số trứng bán ngày thứ ba là:  (\(\dfrac{1}{4}x+224\)) x \(\dfrac{1}{5}\) = \(\dfrac{x+896}{20}\) (quả trứng)

Theo bài ra ta có phương trình:

\(x\) x \(\dfrac{1}{4}\) + 224 + \(\dfrac{x+896}{20}\) = \(x\) 

5\(x\) + 4480 + \(x\) + 896 = 20\(x\)

6\(x\) + 5376 = 20\(x\)

20\(x\) - 6\(x\) = 5376

  14\(x\) = 5376

      \(x\) = 5376 : 14

     \(x\) = 384

Kết luận người đó đem bán số trứng là: 384 quả

\(\dfrac{3}{13}:\dfrac{-11}{-6}+\dfrac{-3}{13}:\dfrac{11}{-5}-\dfrac{2}{13}\)

\(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}\)

\(=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)

Đây là lớp 6 á

2.22 + 3.23 + 4.24 + ... + n.2n  = 2n+11

Đặt vế trái là A ta có:

    A = 2.22 + 3.23 + 4.24 + ... (n -1).2n-1+ n.2n 

2A =  2.23 + 3.24 + 4.25 +....+ (n- 1).2n + n.2n+1

2A-A = [2.23+3.24 + 4.25 +...+(n-1).2n+n.2n+1] - [2.22 + 3.23+...+n.2n]

A   = -2.22+ (2.33-3.23) + (3.24 - 4.24) +...+ [(n-1).2n - n.2n ] + n.2n+1

A = -2.22 - 23 - 24 -...- 2n + n.2n+1

Đặt   B = -2.22 - 23 - 24 - ... - 2n 

     2B  = -2.23 - 24 - 25 -...-2n+1

2B - B = (-2.23 - 24 - 25 -..-2n+1) - (-2.22-23-24-..-2n)

B = -24 -24 - 25 -..2n-2n+1 + 23 + 23 + 24+ 25+ 2n

B = (-24 + 23) + (- 2n+1 + 23) +(-24+24)+(-25+25)+(-2n+2n)

B = -16 + 8 - 2n+1 + 8

B = (-16 + 8 + 8 ) - 2n+1

B = - 2n+1

A = n.2n+1 - 2n+1

Theo bài ra ta có: 

n.2n+1 - 2n+1 = 2n+11

n.2n+1 - 2n+1 - 2n+11 = 0

2n+1.(n - 1 - 210) = 0

Vì n là số tự nhiên nên 2n+1 > 0

Vậy 2n+1.(n - 1- 210) = 0 ⇔ n - 1 - 210 = 0 ⇒ n = 1 + 210 ⇒ n =  1025

Vậy n = 1025

Bạn tặng GP cho mình nhé !(^ ^)

B = (1 - \(\dfrac{1}{2^2}\)).(1 - \(\dfrac{1}{3^2}\)).(1 - \(\dfrac{1}{4^2}\))...(1 - \(\dfrac{1}{201^2}\))

B = \(\dfrac{2^2-1}{2^2}\).\(\dfrac{3^2-1}{3^2}\).\(\dfrac{4^2-1}{4^2}\)...\(\dfrac{201^2-1}{201^2}\)
B = \(\dfrac{4-1}{2^2}\).\(\dfrac{9-1}{3^2}\).\(\dfrac{16-1}{4^2}\)...\(\dfrac{40401-1}{201^2}\)

B = \(\dfrac{3}{2^2}\).\(\dfrac{8}{3^2}\).\(\dfrac{15}{4^2}\)....\(\dfrac{40400}{201^2}\)

B = \(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\).\(\dfrac{3.5}{4.4}\).\(\dfrac{4.6}{5.5}\)...\(\dfrac{200.202}{201.201}\)

B = \(\dfrac{1}{2}\).\(\dfrac{202}{201}\)

B = \(\dfrac{101}{201}\)

$1-\frac{63}{64}=\frac{1}{64}$

$1-\frac{32}{33}=\frac{1}{33}$

Vì $\frac{1}{64}<\frac{1}{33}\Rightarrow \frac{63}{64}>\frac{32}{33}$

Vậy $\frac{63}{64}>\frac{32}{33}$

$\frac{2020}{2019}-1=\frac{1}{2019}$

$\frac{2022}{2021}-1=\frac{1}{2021}$

Mà $\frac{1}{2019}>\frac{1}{2021}$

$\Rightarrow \frac{2020}{2019}>\frac{2022}{2021}$

Vậy $\frac{2020}{2019}>\frac{2022}{2021}$

Lời giải:

a.

$\frac{8}{23}.\frac{46}{24}-\frac{2}{5}x=\frac{1}{3}$

$\frac{2}{3}-\frac{2}{5}x=\frac{1}{3}$

$\frac{2}{5}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$
$x=\frac{1}{3}: \frac{2}{5}=\frac{5}{6}$
b.

$\frac{10}{12}: \frac{2}{3}x=\frac{28}{9}.\frac{3}{56}$

$\frac{5}{4}x=\frac{1}{6}$

$x=\frac{1}{6}: \frac{5}{4}=\frac{2}{15}$

c.

$\frac{x-1}{24}=\frac{2}{x+1}$
$(x-1)(x+1)=2.24$

$x^2-1=48$

$x^2=49=7^2=(-7)^2$
$\Rightarrow x=7$ hoặc $x=-7$
d.

$(\frac{3}{4}x+\frac{1}{4}-\frac{1}{3}): (2+\frac{1}{6}-\frac{1}{4})=\frac{7}{46}$

$(\frac{3}{4}x-\frac{1}{12}):\frac{23}{12}=\frac{7}{46}$

$\frac{3}{4}x-\frac{1}{12}=\frac{7}{46}.\frac{23}{12}=\frac{7}{24}$

$\frac{3}{4}x=\frac{7}{24}+\frac{1}{12}=\frac{3}{8}$

$x=\frac{3}{8}: \frac{3}{4}=\frac{1}{2}$

e.

$2\frac{1}{2}x+0,5x=2\frac{1}{4}$

$2,5x+0,5x=2,25$

$x(2,5+0,5)=2,25$

$3x=2,25$

$x=2,25:3=0,75$

f.

$\frac{1}{3}x+\frac{2}{5}(x-1)=0$

$\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0$

$x(\frac{1}{3}+\frac{2}{5})=\frac{2}{5}$

$x.\frac{11}{15}=\frac{2}{5}$

$x=\frac{2}{5}: \frac{11}{15}=\frac{6}{11}$

g.

$x-3\frac{1}{2}x=-2\frac{6}{7}$
$x(1-3\frac{1}{2})=\frac{-20}{7}$

$x.\frac{-5}{2}=\frac{-20}{7}$

$x=\frac{-20}{7}: \frac{-5}{2}=\frac{8}{7}$

h.

$2(\frac{1}{2}x-\frac{1}{3})-\frac{3}{2}=\frac{1}{4}$

$2(\frac{1}{2}x-\frac{1}{3})=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}$
$\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}:2=\frac{7}{8}$

$\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}$

$x=\frac{29}{24}: \frac{1}{2}=\frac{29}{12}$

i.

$-2\frac{1}{3}x+1\frac{3}{4}x+3\frac{2}{3}=3\frac{1}{2}$

$x(-2\frac{1}{3}+1\frac{3}{4})=3\frac{1}{2}-3\frac{2}{3}$

$x.\frac{-7}{12}=\frac{-1}{6}$

$x=\frac{-1}{6}: \frac{-7}{12}=\frac{2}{7}$

\(\dfrac{4}{15}\) < \(\dfrac{x}{45}\) < \(\dfrac{1}{3}\)

\(\dfrac{4}{15}\) x 45 < \(x\) < \(\dfrac{1}{3}\) x 45

 12 < \(x\) < 15 

** Bổ sung điều kiện $x$ là số tự nhiên.

Lời giải:

$\frac{4}{15}< \frac{x}{45}$
$\Rightarrow \frac{12}{45}< \frac{x}{45}\Rightarrow x> 12(1)$

$\frac{x}{45}< \frac{1}{3}$
$\Rightarrow x< 45.\frac{1}{3}=15(2)$

Từ $(1); (2)\Rightarrow 12< x< 15$

$\Rightarrow x\in \left\{13; 14\right\}$