Vì 13 là lẻ \(\Rightarrow\) 13, 13², 13³, 13⁴, 13⁵, 13⁶ là lẻ.

Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶ là chẵn. \(\Rightarrow\) 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶ \(⋮\) 2

\(\Rightarrow\) ĐPCM

a: =>2x+7/2=16/3:8/3=2

=>2x=-3/2

hay x=-3/4

b: =>8/3x=3+1/3+8+2/3=12

=>x=12:8/3=12x3/8=36/8=9/2

c: =>2x=-2/13

hay x=-1/13

d)

đặt A = 1 + 2 + 2² + ... + 2⁸⁰ 

2A = 2 + 2² + 2³ + ... + 2⁸¹

2A - A = ( 2 + 2² + 2³ + ... + 2⁸¹ ) - ( 1 + 2 + 2² + ... + 2⁸⁰ )

A = 2⁸¹ - 1 > 2⁸¹ - 2

e) 

đặt \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{899}{900}\)

\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{900}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)

\(A=29-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)

đặt \(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\)

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)

\(=1-\frac{1}{30}=\frac{29}{30}< 1\)

\(\Rightarrow A< 29\)

So sánh C và D biết
C=1+13+13^2+...+13^13/1+13+13^2+...+13^12
D=1+11+11^2+...+11^13/1+11+11^2+...+11^12

dễ

bai de ma

Bài 1 :

2⁶ : 2⁴ + 3² - 2³ 

= 2^6-4 + 9 - 8

= 2² + 9 - 8 

= 4 + 9 - 8 

= 13 - 8 

= 5 

2¹⁵ : 2¹⁴ + 3¹⁴ : 3¹³ 

= 2^15-14 + 3^14-13

= 2¹ + 3¹ 

= 2 + 3 

 = 5

a, x=\(\frac{7}{30}\)