a,(2x+1)(y-3)=12

2x+1 và y-3 Ư(12)=

=>x=0,y=15

c) Ta có: \(36^{25}=\left(6^2\right)^{25}=6^{50}\)

\(25^{36}=\left(5^2\right)^{36}=5^{72}\)

Ta có: \(6^{50}=\left(6^5\right)^{10}=7776^{10}\)

mà \(5^{70}=\left(5^7\right)^{10}=78125^{10}\)

nên \(6^{50}< 5^{70}\)

mà \(5^{70}< 5^{72}\)

nên \(6^{50}< 5^{72}\)

hay \(36^{25}< 25^{36}\)

a/

Với $x,y$ là số tự nhiên $2x+1, y-3$ là số nguyên. Mà $(2x+1)(y-3)=12$ nên $2x+1$ là ước của 12. 

$2x+1>0, 2x+1$ lẻ nên $2x+1\in \left\{1;3\right\}$

Nếu $2x+1=1\Rightarrow y-3=12$

$\Rightarrow x=0; y=15$

Nếu $2x+1=3\Rightarrow y-3=4$

$\Rightarrow x=1; y=7$ 

Vậy...........

b/

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8$

$2^x(1+2+2^2+2^3+...+2^{2015})=2^{2019}-8(1)$
$2^x(2+2^2+2^3+2^4+...+2^{2016})=2^{2020}-16(2)$ (nhân 2 vế với 2)

Lấy (2) trừ (1) theo vế thì:

$2^x(2^{2016}-1)=2^{2020}-2^{2019}-8$

$2^x(2^{2016}-1)=2^{2019}(2-1)-8=2^{2019}-8$

$2^x(2^{2016}-1)=2^3(2^{2016}-1)$

$\Rightarrow 2^x=2^3$

$\Rightarrow x=3$

c/

$25^{36}=(5^2)^{36}=5^{72}$

$36^{25}=(6^2)^{25}=6^{50}=(6^5)^{10}< (5^7)^{10}=5^{70}< 5^{72}$

$\Rightarrow 25^{36}> 36^{25}$

\(2VT=2^{x+1}+2^{x+2}+2^{x+3}+...+...+2^{x+2016}\)

\(VT=2VT-VT=2^{x+2016}-2^x=2^{2016}.2^x+2^x=2^x\left(2^{2016}+1\right)\)

\(VP=2^{2019}-2^3=2^3\left(2^{2016}-1\right)\)

\(\Rightarrow2^2\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)

\(\Rightarrow2^x=2^3\Rightarrow x=3\)

\(2^x+2^{x+1}+2^{x+2}+2^{x+2015}=2^{2019}-8\left(1\right)\)

Đặt \(S=2^x+2^{x+1}+2^{x+2}+2^{x+2015}\)

\(\Rightarrow S+\left(1+2^2+...2^{x-1}\right)=\left(1+2^2+...2^{x-1}\right)+2^x+2^{x+1}+2^{x+2}+2^{x+2015}\)

\(\Rightarrow S+\dfrac{2^{x-1+1}-1}{2-1}=1+2^2+...2^{x-1}+2^x+2^{x+1}+2^{x+2}+2^{x+2015}\)

\(\Rightarrow S+2^x-1=\dfrac{2^{x+2015+1}-1}{2-1}\)

\(\Rightarrow S+2^x-1=2^{x+2016}-1\)

\(\Rightarrow S=2^{x+2016}-2^x\)

\(\left(1\right)\Rightarrow2^{x+2016}-2^x=2^{2019}-8=2^{2019}-2^3\)

\(\Rightarrow2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)

\(\Rightarrow2^x=2^3\Rightarrow x=3\)

Kiểm tra lại đề câu b nhé!

phân tích rồi bấm máy nha bạn !

cho 4 nữa cho tròn

(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0

=> 2x + 2x + ... + 2x + (2015 + 1).2015 : 2 = 0

=> 2015.2x + 2031120 = 0

=> 4030x = -2031120

=> x = -504

em mới hc lp 5 thui ạ! cho em xin lỗi nha!

(2x + 1) + (2x + 2) + ... + (2x + 2015) = 0

=> 2x + 2x + ... + 2x + (2015 + 1).2015 : 2 = 0

=> 2015.2x + 2031120 = 0

=> 4030x = -2031120

x = { 0; 1; 2; 3; .......; n } ( n ∈ N )

( 2x+1)+ (2x+2) +...+ ( 2x+2015)

=2x+1+ 2x+2+...+ 2x+2015

=2015.2x +( 1+2+3+...+2015 )

= 4130x + (2015+1). 2015 : 2

=4130x + 2031120

suy ra: 4130x = -2031120

                  x = -491,7966...

= bao nhiu ha bn?