1/41>3/124>5/207>2/83

Ta co

3/124=30/1240

1/41=30/1230

5/207=30/1242

2/83=30/1245

k cho mik nha!

Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\)N*)

Ta có:

\(A=\frac{3^{123}+1}{3^{125}+1}< \frac{3^{123}+1+2}{3^{125}+1+2}\)

\(A< \frac{3^{123}+3}{3^{125}+3}\)

\(A< \frac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\)

\(A< \frac{3^{122}+1}{3^{124}+1}=B\)

=> A < B

\(9A=\frac{3^{125}+9}{3^{125}+1}=1+\frac{8}{3^{125}+1}\)

\(9B=\frac{3^{124}+9}{3^{124}+1}=1+\frac{8}{3^{124}+1}\)

Mà 3^125+1>3^124+1         =>\(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)

Nên A<B

9A=\(\frac{3^{125}+9}{3^{125}+1}\)=\(1+\frac{8}{3^{125}+1}\)

9B=\(\frac{3^{124}+9}{3^{124}+1}\)=\(1+\frac{8}{3^{124}+1}\)

Vì \(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)\(\Rightarrow9B>9A\)\(\Rightarrow B>A\)

Vậy B>A

\(\frac{73}{127}

nguyen thanh tung: dê già, dê cụ, biến thái, mất dạy, 

Đồ là sự lỗi lầm của tạo hóa khi sinh ra một đứa không có giáo dục.

2                                                                

2                            

2 nha bạn

\(B=\frac{3^{122}}{3^{124}+1}=\frac{3^{123}}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+1}=A\)

Do đó \(A>B\).

\(-\frac{11}{25}>-\frac{25}{25}\) ; \(-\frac{25}{25}>-\frac{25}{124}\)

Suy ra \(-\frac{11}{25}>-\frac{25}{124}\)

11/-25 =-1364/3100

-25/124=-625/3100

do -1364/3100<-625/3100 suy ra 11/-25<-25/124

Ta có:

\(\frac{124124}{125125}=\frac{124}{125}\)

Ta so sánh 2 phân số \(\frac{123}{124}và\frac{124}{125}\)

xét chữ số tận cùng khi quy đồng 2 phân số

123x125=.....5

124x124=......6

=> 123x125 < 124x124 

Vậy \(\frac{123}{124}< \frac{124124}{125125}\)

Ai thấy đúng thì nhấn vào chữ " đúng" cho mik nha thạnk you các bạn nhiều

Ta có:

\(\frac{124124}{125125}=\frac{124}{125}\)

Ta so sánh hai phân số \(\frac{123}{124}\)và \(\frac{124}{125}\)

\(1-\frac{123}{124}=\frac{1}{124}\)

\(1-\frac{124}{125}=\frac{1}{125}\)

Vì \(\frac{1}{124}>\frac{1}{125}\)nên suy ra \(\frac{123}{124}< \frac{124}{125}\)

A = \(\dfrac{3^{123}+1}{3^{125}+1}\)  Vì 3¹²³ + 1 < 2¹²⁵ + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)

A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B

Vậy A < B 

\(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)

      \(=\frac{124}{1984}.\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+...+\frac{1}{16}-\frac{1}{2000}\right)\)

       \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

  Và \(B=\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\)

             \(=\frac{1}{16}\left[\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\right]\)

              \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right)\right]\)

=      \(\frac{1}{16}\)  .    \(\left[\left(1+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{1984}-\frac{1}{17}-...-\frac{1}{1984}\right)-\left(\frac{1}{1985}+...+\frac{1}{2000}\right)\right]\)

 = \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

Vậy A = B