Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi

2a.

$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$

$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)

2b.

Gọi tổng trên là $T$

Chứng minh vế đầu tiên:

Ta có:

$\frac{1}{101}> \frac{1}{150}$

$\frac{1}{102}> \frac{1}{150}$

....

$\frac{1}{149}> \frac{1}{150}$

$\Rightarrow T> \underbrace{\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}}_{50}=\frac{50}{150}=\frac{1}{3}$ (đpcm)

Chứng minh vế số 2:

$\frac{1}{101}< \frac{1}{100}$

$\frac{1}{102}< \frac{1}{100}$

....

$\frac{1}{150}< \frac{1}{100}$

$\Rightarrow T< \underbrace{\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}}_{50}=\frac{50}{100}=\frac{1}{2}$ (đpcm)

c) P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

Dễ thấy \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}.50=\dfrac{1}{3}\)(1)

Tương tự

 \(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>50.\dfrac{1}{200}=\dfrac{1}{4}\)(2) 

Từ (1) và (2) ta được

\(P>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) 

P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

         \(\overline{50\text{ hạng tử }}\)                            \(\overline{50\text{ hạng tử }}\)

\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\) 

\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Rightarrow P< \dfrac{5}{6}< 1\)

\(=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{50\cdot\dfrac{49}{2}}\)

\(=\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{50\cdot\dfrac{49}{2}}\)

\(=\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{49\cdot50}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

=2*24/50=48/50=24/25

refer

https://hoc247.net/hoi-dap/toan-6/chung-minh-1-101-1-102-1-103-1-104-1-299-1-300-2-3-faq302038.html

thamkhao

https://hoc247.net/hoi-dap/toan-6/chung-minh-1-101-1-102-1-103-1-104-1-299-1-300-2-3-faq302038.html

9092 = 0

cái này + mỗi phân số vs 1 á 

gặp mấy dạng này + hoặc - cho 1 số nào đó là giải đc , bn tự lm xem

a: =>1/3(2x-5)=-2/3-3/2=-4/6-9/6=-13/6

=>2x-5=-13/6*3=-13/2

=>2x=-3/2

=>x=-3/4

b: =>2/5x=-3/4-1/2=-5/4

=>x=-5/4:2/5=-5/4*5/2=-25/8

a)

 \(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{13}{6}\\ \Rightarrow2x-5=-\dfrac{13}{6}:\dfrac{1}{3}=-\dfrac{13}{2}\\ \Rightarrow2x=-\dfrac{13}{2}+5\\ \Rightarrow2x=-\dfrac{3}{2}\\ \Rightarrow x=-\dfrac{3}{2}:2\\ \Rightarrow x=-\dfrac{3}{4}\)

b)

\(\dfrac{2}{5}x+\dfrac{1}{2}=-\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{3}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{5}{4}:\dfrac{2}{5}=-\dfrac{25}{8}\)

`#040911`

`a)`

`3 1/3 x + 16 3/4 = -13,25`

`=> 3 1/3 x = -13,25 - 16 3/4`

`=> 3 1/3 x = -30`

`=> x = -30 \div 3 1/3`

`=> x =-9`

Vậy, `x = -9`

`b)`

`3 2/7*x - 1/8 = 2 3/4`

`=> 3 2/7x = 2 3/4 + 1/8`

`=> 3 2/7x = 23/8`

`=> x = 23/8 \div 3 2/7`

`=> x = 7/8`

Vậy, `x = 7/8`

`c)`

`x \div 4 1/3 = -2,5`

`=> x = -2,5 * 4 1/3`

`=> x = -65/6`

Vậy, `x = -65/6`

`d)`

`( (3x)/7 + 1) \div (-4) = (-1)/28`

`=> (3x)/7 +1 = (-1)/28 * (-4)`

`=> (3x)/7 + 1 = 1/7`

`=> (3x)/7 = 1/7 - 1`

`=> (3x)/7 = -6/7`

`=> 3x = -6`

`=> x = -6 \div 3`

`=> x = -2`

Vậy, `x = -2.`

a

=>10/3 . x + 16 + 3/4 = -13,25

=>10/3 x + 3/4 = -29,25

=>10/3 x = -30

=>x=-30 : 10/3

=>x=-30 . 3/10

=>x=-9

b.

=>23/7 x - 1/8 = = 11/4

=>23/7 x = 11/4 + 1/8

=>23/7 x= 22/8 + 1/8

=>23/7 x= 23/8

=>x=23/8 : 23/7

=>x=23/8 . 7/23

=>x=7/8

c.

=>x : 13/3 =-5/2

=>x=-5/2 . 13/3

=>x=-65/6

d.

=>3x/7 +1 = (-1/28) . (-4)

=>3x/7 + 1 = 1/7

=>3x/7 = -6/7

=>3x=-6

=>x=-2

a) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)

\(\Rightarrow\dfrac{10}{3}x+\dfrac{67}{4}=-\dfrac{53}{4}\)

\(\Rightarrow\dfrac{10}{3}x=-30\)

\(\Rightarrow x=-30:\dfrac{10}{3}\)

\(\Rightarrow x=-9\)

b) \(3\dfrac{2}{7}x-\dfrac{1}{8}=2\dfrac{3}{4}\)

\(\Rightarrow\dfrac{23}{7}x-\dfrac{1}{8}=\dfrac{11}{4}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{11}{4}+\dfrac{1}{8}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{23}{8}\)

\(\Rightarrow x=\dfrac{23}{8}:\dfrac{23}{7}\)

\(\Rightarrow x=\dfrac{7}{8}\)

c) \(x:4\dfrac{1}{3}=-2,5\)

\(\Rightarrow x:\dfrac{13}{3}=-\dfrac{5}{2}\)

\(\Rightarrow x=-\dfrac{5}{2}\cdot\dfrac{13}{3}\)

\(\Rightarrow x=-\dfrac{65}{6}\)

d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{-1}{28}\cdot-4\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\Rightarrow\dfrac{3x}{7}=-\dfrac{6}{7}\)

\(\Rightarrow x=-\dfrac{6}{7}:\dfrac{3}{7}\)

\(\Rightarrow x=-2\)

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)