\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

7000⁴⁴⁴⁴ = ........0

6000⁵⁵⁵⁵ = ........0

mà 5555 > 4444 

=> 7000⁴⁴⁴⁴ < 6000⁵⁵⁵⁵ 

7000⁴⁴⁴⁴ = 1000⁴⁴⁴⁴. 7⁴⁴⁴⁴ 

6000⁵⁵⁵⁵ = 1000⁴⁴⁴⁴.1000¹¹¹¹ . 6⁵⁵⁵⁵ 

=> 7⁴⁴⁴⁴ < 1000¹¹¹¹. 6⁵⁵⁵⁵

=> 7000⁴⁴⁴⁴ < 6000⁵⁵⁵⁵

a, Đề có vẻ sai sai nhé :v

b, \(\left|\frac{1}{2}x-\frac{2}{3}\right|-1=\frac{1}{6}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{2}{3}\right|=\frac{7}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{2}{3}=\frac{7}{6}\\\frac{1}{2}x-\frac{2}{3}=-\frac{7}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{3}\\x=-1\end{cases}}\)

Vậy : ....

c, \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{4}{5}\)

\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\cdot(x+1)}=\frac{4}{5}\)

\(\Leftrightarrow2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{4}{5}\)

\(\Leftrightarrow2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{4}{5}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)

\(\Leftrightarrow x+1=10\Leftrightarrow x=9\)

Vậy x = 9

a) x=0 nhé

\(\frac{3}{4}-\left[\frac{-5}{3}-\left(\frac{1}{2}+\frac{2}{9}\right)\right]\)

\(=\frac{3}{4}-\left[\frac{-5}{3}-\frac{1}{2}-\frac{2}{9}\right]\)

\(=\frac{3}{4}+\frac{5}{3}+\frac{1}{2}+\frac{2}{9}\)

\(=\left(\frac{3}{4}+\frac{1}{2}\right)+\left(\frac{5}{3}+\frac{2}{9}\right)\)

\(=\frac{5}{4}+\frac{17}{9}\)

\(=\frac{113}{36}\)

=))

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

   = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

   = \(1-\frac{1}{20}\)

   = \(\frac{19}{20}\)

Vậy A = \(\frac{19}{20}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(A=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)

Vậy A = 19/20

Vì sao lại bé hơn?

Vì nếu là số bé mà lũy thừa lớn thì số đó sẽ lớn hơn số lớn mà lũy thừa bé

Hc tốt

#)Giải :

\(\left(x-2\right)^4=\left(x-2\right)^6\)

\(\Rightarrow x-2\in\left\{-1;0;1\right\}\)

\(\Rightarrow x\in\left\{1;2;3\right\}\)

\(2x+2x=40\)

\(\Rightarrow2x=20\)

\(\Rightarrow x=10\)

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\left(1-\frac{1}{2020}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}\)

Số nào xuất hiện 2 lần thì thay thế những số đó bằng số 1.

\(B=\frac{1}{2020}\)

B = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2019}\right).\left(1-\frac{1}{2020}\right)\)

    = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2018}{2019}.\frac{2019}{2020}\)

    = \(\frac{1.2.3...2019}{2.3.4..2020}\)(Nếu có 2 thừa số giống nhau lặp lại ở tử số và mẫu số thì rút gọn coi như triệt tiêu hết và không có gì)

   =  \(\frac{1}{2020}\)