\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\) 

\(\Rightarrow\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) 

\(\Rightarrow\dfrac{4^5.4.6^5.6}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{\left(2.2\right)^5.2.2.\left(3.2\right)^5.3.2}{3^5.3.2^5.2}=2^n\) 

\(\Rightarrow\dfrac{2^5.2^5.2.2.3^5.2^5.3.2}{3^5.3.2^5.2}=2^n\) 

Rút gọn vế trái ta có :

\(2^5.2.2.^5=2^n\)

\(\Rightarrow2^{12}=2^n\) 

\(\Rightarrow n=12\) ( Thỏa mãn điều kiện \(n\in N\) ) 

Vậy n =12 

Câu hỏi của Lê Khánh Nhi - Toán lớp 7 - Học toán với OnlineMath sửa n thành x cho sửa cho nó thành lũy thừa luôn

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)

<=>\(\dfrac{4.4^5}{3.3^5}\cdot\dfrac{6.6^5}{2.2^5}=2^n\)

<=>\(\dfrac{4^6.6^6}{3^6.2^6}\)=2ⁿ

<=>\(\dfrac{\left(4.6\right)^6}{\left(3.2\right)^6}=2^n\)

<=>4⁶=2ⁿ

<=>(2²)⁶=2ⁿ

<=>2ⁿ=2¹²

<=>n=12

Sửa đề: 3^5+3^5+3^5; 2^x

=>\(2^x=\dfrac{4^5\cdot4}{3^5\cdot3}\cdot\dfrac{6^5\cdot6}{2^5\cdot2}\)

=>\(2^x=\left(\dfrac{4}{3}\right)^6\cdot\left(\dfrac{6}{2}\right)^6=4^6=2^{12}\)

=>x=12

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\dfrac{4.4^5.6.6^5}{3.3^5.2.2^5}\)

\(=\dfrac{4^6.6^6}{3^6.2^6}\)

\(=\dfrac{2^6.2^6.2^6.3^6}{3^6.2^6}\)

\(=2^{12}=2^{3^4}=8^4=8^x\)

Vậy x = 4

Vậy x = 4

Lời giải:

\(\text{VT}=\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\frac{4.4^5}{3.3^5}.\frac{6.6^5}{2.2^5}=\frac{4^6.6^6}{3^6.2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)

Do đó: \(8^{|2x+6|}=2^{12}\Leftrightarrow 2^{3|2x+6|}=2^{12}\)

\(\Leftrightarrow 3|2x+6|=12\Leftrightarrow |2x+6|=4\)

\(\Rightarrow\left[{}\begin{matrix}2x+6=4\\2x+6=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

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