Xin sửa lại đề : \(\left(2x-11\right)^5=\left(2x-11\right)^3\)

\(\left(2x-11\right)^5=\left(2x-11\right)^3.\)

\(\Rightarrow\orbr{\begin{cases}2x-11=0\\2x-11=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=6\end{cases}}}\)

giair giúp mình với

Làm thử nha .

\(a,\)Ta có :

\(789^{123}=789^{100}.789^{23}\)

\(1000^{10}=10^{3^{10}}=10^{30}\)

Vì \(789^{100}>10^{30}\)( thấy rõ )

=> 789¹⁰⁰.789²³ > 10³⁰

Hay \(789^{123}>1000^{10}\)

789¹²³>1000¹⁰

124⁸⁷⁵<999⁸⁸⁸

^{tt nnha}

\(\frac{4n+25}{n+2}=\frac{4n+8+17}{n+2}=\frac{4.\left(n+2\right)+17}{n+2}=4+\frac{17}{n+2}\)

\(4\in Z\Rightarrow n+2\inƯ\left(17\right)\)

\(\Rightarrow n+2\in\left\{-17;-1;1;17\right\}\)

\(\Rightarrow n\in\left\{-19;-3;-1;15\right\}\)

\(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.71\)

\(72^{44}-72^{43}=72^{43}.\left(72-1\right)=72^{43}.71\)

Vì \(72^{44}>72^{43}\Rightarrow72^{45}-72^{44}>72^{44}-72^{43}\)

a)\(\frac{27-5n}{n}=\frac{27}{n}-\frac{5n}{n}=\frac{27}{n}-5\)

\(\Rightarrow n\inƯ\left(27\right)=\left\{-27;-9;-3;-1;1;3;9\right\}\)

b)\(\frac{n+6}{n+2}=\frac{n+2+4}{n+2}=\frac{n+2}{n+2}+\frac{4}{n+2}=1+\frac{4}{n+2}\)

\(\Rightarrow n+2\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow n\in\left\{-6;-4;-3;-1;0;2\right\}\)

(n+6) chia hết cho (n+2)

=> (n+4+2) chia hết cho (n+2)

=> 4.(n+2) ( do n+2 chia hết cho n+2)

=> n+2 thuộc {1;4}

=> n thuộc {2}

Vậy n thuộc {2}

25 mũ 30 lớn hơn

125 mũ 19 lớn hơn 

Chắc chắn 100%

2^2.8^27>16^25 mk nha

2^2*8^27>16^25 ok

Thank