S-P= (1 - 1/2 + 1/3 - 1/4 +...+ 1/2011 - 1/2012 + 1/2013) - ( 1/1007 + 1/1008 +...+ 1/2012 + 1/2013 )
 S-P= (1- 1/2 + ... + 1/1005 - 1/1006) - 2.(1/1008 + 1/1010 + 1/1012 +...+ 1/2012)
 S-P= 1+1/2+1/3+...+1/1006 - 2.( 1/2 + 1/4 + 1/6 +...+ 1/2012)
 S-P= 1 + 1/2 + 1/3 +...+ 1/1006 - ( 1+ 1/2 + 1/3 +...+ 1/1006 )
 S-P= 0
⇒ (S-P)^2013 = 0

\(=\frac{4^5.\left(3^2\right)^4}{\left(4^2\right)^2.\left(3^3\right)^3}=\frac{4^5.3^8}{4^4.3^9}=\frac{4}{3}\)

\(=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^4\right)^2\cdot\left(3^3\right)^3}\)

\(=\frac{2^{10}\cdot3^8}{2^8\cdot3^9}\)

\(=\frac{4}{3}\)

thêm đk : a,b,c > 0

Ta có :

\(\frac{a}{a+b}< 1\)\(\Rightarrow\frac{a}{a+b+c}< \frac{a}{a+b}< \frac{a+c}{a+b+c}\)( 1 )

\(\frac{b}{b+c}< 1\)\(\Rightarrow\frac{b}{a+b+c}< \frac{b}{b+c}< \frac{a+b}{a+b+c}\)( 2 )

\(\frac{c}{c+a}< 1\)\(\Rightarrow\frac{c}{a+b+c}< \frac{c}{c+a}< \frac{b+c}{a+b+c}\)( 3 )

cộng ( 1 ), ( 2 ) và ( 3 ) ta được :

\(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}\)

\(\Leftrightarrow1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)

Vậy M không phải là số nguyên

Có : a/a+b > 0 => a/a+b > a/a+b+c

Tương tự : b/b+c > b/a+b+c ; c/c+a > c/a+b+c

=> M > a+b+c/a+b+c = 1 (1)

Lại có : a < a+b => a/a+b < 1 => 0 < a/a+b < 1 => a/a+b < a+c/a+b+c

Tương tự : b/b+c < b+a/a+b+c ; c/c+a < c+b/a+b+c

=> M < a+c+b+a+c+b/a+b+c = 2 (2)

Từ (1) và (2) => 1 < M < 2

=> M ko phải là số tự nhiên

Tk mk nha

3^2n = (3^2)^n = 9^n

2^3n = (2^3)^n = 8^n

Vì 9^n > 8^n => 3^2n > 2^3n

7.2^13 < 8.2^13 = 2^3.2^13 = 2^3+13 = 2^16

=> 7.2^13 < 2^16

Tk mk nha

bạn Nguyễn Anh Quân bạn nên xen lại câu 7.2¹³ và 2¹⁶ đi bạn

\(\left|\frac{6}{7}x-4\right|< \frac{2}{7}\)

\(\Leftrightarrow-\frac{2}{7}< \frac{5}{7}x-4< \frac{2}{7}\)

\(\Leftrightarrow5\frac{1}{5}< x< 6\)

=> -2/7 < 5/7.x-4 < 2/7

=> -2/7+4 < 5/7.x < 2/7+4

=> 26/7 < 5/7.x < 30/7

=> 26/7 : 5/7 < x < 30/7 : 5/7

=> 26/5 < x < 6

Tk mk nha

A = 1.2 + 2.3 + 3.4 + … + n.(n + 1)

3A= 1.2.3 + 2.3.3 + 3.4.3 + ... + n.(n+1).3

3A = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + ... + n.(n+1).(n+2-n+1)

3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + n.(n+1).(n+2) - (n-1).n.(n+1)

3A = n.(n+1).(n+2) 

A = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Ta có : 3A = 1.2.3 + 2.3.3 + … + n(n + 1).3 = 1.2.(3 - 0) + 2.3.(3 - 1) + … + n(n + 1)[(n - 2) - (n - 1)] = 1.2.3 - 1.2.0 + 2.3.3 - 1.2.3 + … + n(n + 1)(n + 2) - (n - 1)n(n + 1) = n(n + 1)(n + 2) 

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

\(=3^n\cdot3^2+3^n-2^n.2^2-2^n\)

\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)