ai trả lời được thì tôi k đúng cho :))

3^-2 . 3⁴ . 3^x = 3⁷

<=> 3^{2 + x} = 3⁷

<=> 2 + x = 7

<=> x = 7 - 2

<=> x = 5

\(2^{-2}\cdot2^x+2\cdot2^x=9\cdot2^6\\ \Rightarrow2^{x-2}+2^{x+1}=9\cdot2^6\\ \Rightarrow2^{x-2}\left(1+2^3\right)=9\cdot2^6\\ \Rightarrow2^{x-2}\cdot9=9\cdot2^6\Rightarrow2^{x-2}=2^6\\ \Rightarrow x-2=6\Rightarrow x=8\)

\(3^{-2}\cdot3^4\cdot3^x=3^7\\ \Rightarrow3^{x+4-2}=3^7\\ \Rightarrow x+2=7\Rightarrow x=5\)

1) \(2^{-2}.2^x+2.2^x=9.2^6\)

\(\Rightarrow2^x.\left(2^{-2}+2\right)=576\)

\(\Rightarrow2^x.\dfrac{9}{4}=576\)

\(\Rightarrow2^x=256=2^8\Rightarrow x=8\)

2) \(\Rightarrow3^{x+2}=3^7\Rightarrow x+2=7\Rightarrow x=5\)

Ta có : 8 . 6 + 288 : (x - 3)² = 50

=> 48 + 288 : (x - 3)² = 50

=> 288 : (x - 3)² = 50 - 48

=> 288 : (x - 3)² = 2

=> (x - 3)² = 288 : 2

=> (x - 3)² = 144

=> không tìm được x thỏa mãn điều kiện trên

8.6+288:(x-3)² = 50

<=> 48+288:(x-3)² = 50

<=> 288:(x-3)² = 50-48 = 2

<=> (x-3)² = 2.288 = 576 = 24²

<=> \(\orbr{\begin{cases}x-3=24\\x-3=-24\end{cases}}\)

<=> \(\orbr{\begin{cases}x=24+3=27\\x=3-24=-21\end{cases}}\)

Vậy x thuộc { 27 ; -21 }

\(8\cdot6+288:\left(x-3\right)^2=50\)

\(48+288:\left(x-3\right)^2=50\)

\(288:\left(x-3\right)^2=50-48\)

\(288:\left(x-3\right)^2=2\)

\(\left(x-3\right)^2=288:2\)

\(\left(x-3\right)^2=144\)

\(Th1:x-3=12\)\(Th2:x-3=-12\)

\(x=12+3=15\)\(x=-12+3=-9\)

Vậy \(x=15\)hoặc \(x=-9\)

\(6^2.2^2.5:\left[3.12-\left(2x-6\right)\right]=2^3.5\)

\(\Rightarrow36.4.5:\left[36-\left(2x-6\right)\right]=8.5\)

\(\Rightarrow720:\left[36-\left(2x-6\right)\right]=40\)

\(\Rightarrow36-\left(2x-6\right)=720:40=18\)

\(\Rightarrow2x-6=36-18=18\)

\(\Rightarrow2x=18+6=24\)

\(\Rightarrow x=24:2=12\)

8.6+288:(x-3)^2=50

48+288:(x-3)^2=50

288:(x-3)^2=50-48

288:(x-3)^2=2

(x-3)^2=288:2

(x-3)^2=144

(x-3)^2=12^2

x-3=12

x=12+3

x=15

Vậy x=15

HT

\(8\cdot6+288:\left(x-3\right)^2=50\)

\(48+288:\left(x-3\right)^2=50\)

\(288:\left(x-3\right)^2=50-48=2\)

\(\left(x-3\right)^2=288:2=144\)

Vì x là số tự nhiên nên (x-3) là số tự nhiên.

\(\left(x-3\right)^2=12^2\)

\(x-3=12\)

\(x=12+3=15\)

\(\Leftrightarrow\dfrac{3\cdot2^{x+1}}{2^{x+3}}=40\\ \Leftrightarrow\dfrac{3}{2^2}=40\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)