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/

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

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

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

ukm xong rồi

(x-1)²⁰²⁰=(x-1)²⁰²²

=>(x-1)²⁰²⁰-(x-1)²⁰²²=0

=>(x-1)²⁰²⁰-(x-1)²⁰²⁰.(x-1)²=0

=>(x-1)²⁰²⁰(1-(x-1)²=0

=>(x-1)²⁰²⁰=0 hoặc 1-(x-1)²=0

=>x=1 hoặc x=2.

Bài 2

a,2¹⁰⁵ và 5⁴⁵

2¹⁰⁵=(2⁷)¹⁵=128¹⁵

5⁴⁵=(5³)¹⁵=125¹⁵

Vì 128^15_>12515 nên 2¹⁰⁵>5⁴⁵.

b,

5⁵⁴ và 3⁸¹

5⁵⁴=(5⁶)⁹=15625⁹

3⁸¹=(3⁹)⁹=19683⁹

Vì 15625⁹<19683⁹ nên 5⁵⁴<3⁸¹

Bài 1 :

\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)

\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)

\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=1\\x-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

18743+(1409x6-3625)

= 18743+(8454-3625)

= 18743+4829

= 23572

b

B

B

p = 1+ \(\dfrac{x+1}{\sqrt{x}}\) sẽ lớn hơn -1 vì \(\sqrt{x}\) => x dương =>  \(\dfrac{x+1}{\sqrt{x}}\)> 0

Ta có: \(P-1=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\)

\(=\dfrac{x+1}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ

Suy ra: P>1