1 - 1/2 + 1/3 - 1/4 + .. + 1/121 - 1/122 + 1/123

=  ( 1 + 1/3 + ... + 1/121 + 1/123 ) - ( 1/2 + 1/4 + .. + 1/122 ) 

Đến bước này ta sẽ cùng cộng 2 vế với : 1/2 + 1/4 + .. + 1/122 

=   ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/121 + 1/122 + 1/123 ) - 2. ( 1/2 + 1/4 + .. + 1/122 )

=    (  1 + 1/2 + 1/3 + ...+ 1/122 +1 /123 )  -  (  1 + 1/2 + ...+ 1/61 ) 

=                   1/62 + 1/63 + ..+1/122 + 1/123 

Chúc học giỏi !!

Xét \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{123}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{122}\right)\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-2\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{61}\right)\)

\(=\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+...+\frac{1}{123}\)

\(\dfrac{x+1}{124}+1+\dfrac{x+2}{123}+1=\dfrac{x+3}{122}+1+\dfrac{x+4}{121}+1\)

\(\Leftrightarrow\dfrac{x+125}{124}+\dfrac{x+125}{123}=\dfrac{x+125}{122}+\dfrac{x+125}{121}\)

\(\Leftrightarrow\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\ne0\right)=0\Leftrightarrow x=-125\)

<=>\(\dfrac{x+1}{124}+\dfrac{x+2}{123}-\dfrac{x+3}{122}-\dfrac{x+4}{121}=0\)

<=>\(\left(\dfrac{x+1}{124}+1\right)+\left(\dfrac{x+2}{123}+1\right)-\left(\dfrac{x+3}{122}+1\right)-\left(\dfrac{x+4}{121}+1\right)=0\)

<=>\(\dfrac{x+125}{124}+\dfrac{x+125}{123}-\dfrac{x+125}{122}-\dfrac{x+125}{121}=0\)

<=>\(\left(x+125\right)\left(\dfrac{1}{124}+\dfrac{1}{123}-\dfrac{1}{122}-\dfrac{1}{121}\right)=0\)

<=>x+125=0

<=>x=-125

Số số hạng  : (123 - 1) : 1 + 1 = 123

Tổng : (123 + 1) x 123 : 2 = 7626

( 1 + 123 ) . 124 : 2 = 7688

Dãy số có số số hạng là:

    (123-1):1+1=123

Tổng dãy số là;

   (123+1)x123:2=7626

             Đ/S: 7626

Giá trị của X trong phép tính 486 : X = 4 ( dư 2 ) là : X = 121 

(1/2+2/3+3/4+4/5+...+122/123+123/124).(125-5.25)

=(1/2+2/3+3/4+4/5+...+122/123+123/124).(125-125)

=(1/2+2/3+3/4+4/5+...+122/123+123/124).0=0

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