\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-1=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Vậy x = 1 hoặc x=2

( x-1)^x+2 = (x-1)^x+6

( x-1)^x+2 - (x-1)^x+6 =0

( x-1)^x+2 - (x-1)^x+2 . (x-1)4 =0  

( x-1)^x+2 - [1- (x-2)⁴]  =0

TH1: ( x-1)^x+2 =0

x-1 = 0

x= 0+1 

x = 1

TH2: 

1- (x-2)⁴ =0

(x-2)⁴ = 1-0

(x-2)⁴ =1  ( Hai trường hợp vì số mũ là chẵn bạn nhé, mình biểu thị bằng dấu * )

* x-2 = 1         * x-2=-1

x=3                  x=1

Vậy x = { 1; 3}

a/ \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.\left(2^5+1\right)=2^{15}.33\)chia hết cho 33

b/ \(10^9+10^8+10^7=10^7\left(10^2+10+1\right)=10^6.2.5.111=10^6.2.555\)chia hết cho 555

a,\(5^{28}=25^{14}\) Mà 25<26

\(\Rightarrow5^{28}< 26^{14}\)

Mấy câu sau làm tương tự

a) 5²⁸ và 26¹⁴

\(5^{28}=\left(5^2\right)^{14}=25^{14}\)

Vì \(25^{14}< 26^{14}\)nên \(5^{28}< 26^{14}\)

b) 31¹¹ và 17¹⁴

\(31^{11}< 32^{11}=\left(4.8\right)^{11}=4^{11}.8^{11}=2^{22}.8^{11}\)

\(17^{14}>16^{14}=2^{14}.8^{14}=2^{14}.8^3.8^{11}=2^{14}.2^9.8^{11}=2^{23}.8^{11}\)

Ta có : \(2^{23}.8^{11}>2^{22}.8^{11}\), nên \(16^{14}>32^{11}\)

Vậy \(17^{14}>16^{14}>32^{11}>31^{11}\Rightarrow17^{14}>31^{11}\)

\(\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot...\cdot\left(1-\frac{1}{1326}\right)\)

\(=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot...\cdot\frac{1325}{1326}\)

\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot...\cdot\frac{2650}{2652}\)

\(=\frac{5.8}{6.7}\cdot\frac{6.9}{7.8}\cdot\frac{7.10}{8.9}\cdot....\cdot\frac{50.53}{51.52}\)

\(=\frac{5\cdot6\cdot7\cdot.....\cdot50}{6\cdot7\cdot8\cdot....\cdot51}\cdot\frac{8\cdot9\cdot10\cdot....\cdot53}{7\cdot8\cdot9\cdot...\cdot52}=\frac{5}{51}\cdot\frac{53}{7}=\frac{265}{357}\)

\(\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{1326}\right)\)

\(=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot\cdot\cdot\cdot\cdot\frac{1325}{1326}\)

\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot\cdot\cdot\cdot\cdot\cdot\frac{2650}{2652}\)

\(=\frac{5\cdot8}{6\cdot7}\cdot\frac{6\cdot9}{7\cdot8}\cdot\frac{7\cdot10}{8\cdot9}\cdot\cdot\cdot\cdot\cdot\frac{50\cdot53}{51\cdot52}\)

\(=\frac{\left(5\cdot8\right)\left(6\cdot9\right)\left(7\cdot10\right)\cdot\cdot\cdot\cdot\cdot\left(50\cdot53\right)}{\left(6\cdot7\right)\left(7\cdot8\right)\left(8\cdot9\right).....\left(51\cdot52\right)}\)

\(=\frac{\left(5\cdot6\cdot7\cdot\cdot\cdot\cdot\cdot50\right)\cdot\left(8\cdot9\cdot10\cdot\cdot\cdot\cdot\cdot53\right)}{\left(6\cdot7\cdot8\cdot\cdot\cdot\cdot\cdot51\right)\cdot\left(7\cdot8\cdot9\cdot\cdot\cdot\cdot\cdot52\right)}\)

\(=\frac{5\cdot53}{51\cdot7}=\frac{265}{357}\)