có dell câu hỏi ạ

nó bị mất phần câu hỏi rồi xin lỗi mọi người 

bài khó quá bạn ạ

1+1=2

2+2=4

3+3=6

mình 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Thui khỏi bt lm òi * nhắn lun ko bt cho nhanh lại cn lm màu *

\(pt\Leftrightarrow \dfrac{3^x}{3}.\dfrac{4^x}{4}=12^{9-x}\Leftrightarrow 12^{x-1}=12^{9-x}\)

Suy ra x-1=9-x nên x=5

\(3^{x+1}-2^{2x+1}-12^{\dfrac{x}{2}}=0\)

\(\Leftrightarrow3.3^x-2.4^x-\sqrt{3^x.4^x}=0\)

\(\Leftrightarrow3.\dfrac{3^x}{4^x}-\sqrt{\dfrac{3^x}{4^x}}-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{3}{4}\right)^x=-\dfrac{2}{3}\left(l\right)\\\left(\dfrac{3}{4}\right)^x=1\end{matrix}\right.\)

\(\Rightarrow x=0\)

Điều kiện:

So với điều kiện, ta được tập nghiệm của phương trình là  S   =   3 2 ;   3 4 ;   7 8

Vậy phương trình đã cho có 3 nghiệm

Đáp án A

\(2^x.3^{x-1}.5^{x+2}=12\\ \Leftrightarrow2^x.\frac{3^x}{3}.\left(5^x.25\right)=12\\ \Leftrightarrow\left(2.3.5\right)^x=\frac{36}{25}\\ \Leftrightarrow30^x=\frac{36}{25}\\\Leftrightarrow x=log_{30}\left(\frac{36}{25}\right)\)

Bạn chép đề có đúng không? Nếu sửa lại đề bài 1 tý thì nghiệm sẽ đẹp hơn

\(2^x.3^{x-1}.5^{x-2}=12\\ \Leftrightarrow2^x.\frac{3^x}{3}.\frac{5^x}{25}=12\\ \Leftrightarrow\left(2.3.5\right)^x=900\\ \Leftrightarrow30^x=900\Leftrightarrow x=2\)

Cho mk hỏi là làm thế nào để ra 900 ạ