\(\dfrac{2n+1}{n-1}=\dfrac{2n-2+3}{n-1}=\dfrac{2n-2}{n-1}+\dfrac{3}{n-1}=2+\dfrac{3}{n-1}\)

\(\Rightarrow3⋮n-1\Rightarrow n-1\inƯ\left(3\right)\)

\(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)

Xét ước

\(n^2+1⋮n+2\)

\(\Rightarrow n^2+2n-2n+1⋮n+2\)

\(\Rightarrow n^2+2n-2n-4+5⋮n+2\)

\(\Rightarrow n\left(n+2\right)-2\left(n+2\right)+5⋮n+2\)

\(\Rightarrow\left(n-2\right)\left(n+2\right)+5⋮n+2\)

\(\Rightarrow5⋮n+2\)

\(\Rightarrow n+2\inƯ\left(5\right)\)

\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)

Xét ước

\(\dfrac{n^2-3n+2}{n+1}\)

\(\Rightarrow n^2-3n+2⋮n+1\)

\(\Rightarrow n^2+n-4n+2⋮n+1\)

\(\Rightarrow n^2+n-4n-4+6⋮n+1\)

\(\Rightarrow n\left(n+1\right)-4\left(n+1\right)+6⋮n+1\)

\(\Rightarrow\left(n-4\right)\left(n+1\right)+6⋮n+1\)

\(\Rightarrow6⋮n+1\Rightarrow n+1\inƯ\left(6\right)\)

\(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Xét ước

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{2}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(A=1-\frac{1}{10}\)

\(A=\frac{9}{10}\)(đáp án của p sai nha)

= 1 / 1*2 + 1 / 2*3 + 1/ 3*4 + 1 / 4 * 5 ... + 1/ 9*10 = 1-1/2 + 1/2 -1/3 +... + 1/9 - 1/10 = 1 - 1/10 = 9 /10 

đáp án của bạn bị sai rùi

\(A=\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\)

\(\frac{1}{5^2}A=\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+...+\frac{1}{5^{103}}\)

\(\left(1-\frac{1}{5^2}\right)A=\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)-\left(\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+...+\frac{1}{5^{103}}\right)\)

\(\frac{24}{25}A=\frac{1}{5}-\frac{1}{5^{103}}\)

\(A=\left(1-\frac{1}{5^{102}}\right).\frac{5}{24}\)

Suy ra \(\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\div\left(1-\frac{1}{5^{102}}\right)=\frac{5}{24}\).

\(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\)

\(x+\frac{1}{8}=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{1}{8}\)

\(x=\frac{4}{16}-\frac{2}{16}\)

\(x=\frac{1}{8}\)

Vậy \(x=\frac{1}{8}\)

b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\)

      \(\frac{8}{27}-x=\frac{1}{3}\)

                    \(x=\frac{8}{27}-\frac{1}{3}\)

                    \(x=\frac{8}{27}-\frac{9}{27}\)

                     \(x=-\frac{1}{27}\)

Vậy \(x=-\frac{1}{27}\)

c) \(x.\left(-\frac{1}{2}\right)^4=\frac{3}{8}\)

 \(x.\frac{1}{16}=\frac{3}{8}\)

       \(x=\frac{3}{8}:\frac{1}{16}\)

        \(x=\frac{3}{8}.16\)

      \(x=6\)

c) \(\left(\frac{1}{2}\right)^3.x=\left(\frac{1}{2}\right)^5\)

\(x=\left(\frac{1}{2}\right)^5:\left(\frac{1}{2}\right)^3\)

\(x=\left(\frac{1}{2}\right)^2\)

\(x=\frac{1}{4}\)

Vậy \(x=\frac{1}{4}\)

Chúc bạn học tốt !!!

a) \(x+\left(\frac{1}{2}\right)^3=\frac{1}{4}\Leftrightarrow x+\frac{1}{8}=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}-\frac{1}{8}\Leftrightarrow x=\frac{1}{8}\)

b) \(\left(\frac{2}{3}\right)^3-x=\frac{1}{3}\Leftrightarrow\frac{8}{27}-x=\frac{1}{3}\Leftrightarrow-x=\frac{1}{3}-\frac{8}{27}\Leftrightarrow-x=\frac{1}{27}\Leftrightarrow x=-\frac{1}{27}\)

c) \(x.\left(\frac{-1}{2}\right)^4=\frac{3}{8}\Leftrightarrow x.\frac{1}{16}=\frac{3}{8}\Leftrightarrow x=\frac{3}{8}:\frac{1}{16}\Leftrightarrow x=6\)

d) \(\left(\frac{1}{2}\right)^2.x=\left(\frac{1}{2}\right)^5\Leftrightarrow\frac{1}{8}.x=\frac{1}{32}\Leftrightarrow x=\frac{1}{32}:\frac{1}{8}\Leftrightarrow x=\frac{1}{4}\)

yhgfhgfyht

x(x-1) -(x-1) = x(x-1) -1(x-1) = (x-1)(x-1) = (x-1)²

Chịu thua!!!

\(x\left(x-1\right)-\left(x-1\right)\)

\(=x\left(x-1\right)-1\left(x-1\right)\)

\(=\left(x-1\right)\left(x-1\right)\)

Ta có 2009²⁰= 2009^2x10=(2009²)¹⁰= 4036081¹⁰ 

Vì 4036081<20092009
Nên 4036081¹⁰<20092009¹⁰

Vậy...
Rồi đó nha 
~ủng hộ dùm~

Hì hì mik biết có câu 1 thui

a = 1+ 1/2 +1/3+...+1/ 1025 + 1/1026

a= 1+ (1/12+1/3+....+1/1025) - (1/2+1/3+...+1/1025+ 1/1026)

a= 1+ (1/2- 1/1026)

a= 1+ 256/513

a= 283/171

ko chắc chắn

đúng k nha

Ta có:

AB=1+1/2+1/3+...+1/4026/1+1/3+1/5+1/7+...+1/4025

⇒AB=(1+1/3+1/5+...+1/4025)+(1/2+1/4+...+1/2046)1+1/3+1/5+...+1/4025

⇒AB=1+1/3+1/5+...+1/4025/1+1/3+1/5+....+1/4025+1/2+1/4+...+1/4026/1+1/3+1/5+...+1/4025

⇒AB=1+1/2+1/4+...+1/2046/1+1/3+1/5+...+1/4025

Dễ thấy AB>1

Mà 20132014<1