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\(2^{x+2}\cdot3^{x+1}\cdot5x=10800\)
\(2^x\cdot2^2\cdot3^x\cdot2\cdot5^x=10800\)
\(\left(2\cdot3\cdot5\right)^x\cdot12=10800\)
\(30^x=10800\div12\)
\(30^x=900\)
\(\Leftrightarrow30^{x=}=30^2\Leftrightarrow x=2\)
\(\left|x-\dfrac{5}{2}\right|-\dfrac{1}{2}=\dfrac{9}{2}\)
\(\left|x-\dfrac{5}{2}\right|\) \(=\dfrac{9}{2}+\dfrac{1}{2}=\dfrac{10}{2}=5\)
\(\text{Vậy }x-\dfrac{5}{2}=5\)
\(x\) \(=5+\dfrac{5}{2}=\dfrac{15}{2}\)
\(\text{hoặc }x-\dfrac{5}{2}=-5\)
\(x\) \(=\left(-5\right)+\dfrac{5}{2}=\dfrac{-5}{2}\)
\(\Rightarrow x\in\left\{\dfrac{15}{2};\left(\dfrac{-5}{2}\right)\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=5\\x-\dfrac{5}{2}=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Ta có:
\(\left(\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\right)+-\frac{1}{2}=\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\)\(-\frac{1}{2}\)
=\(\frac{6}{30}+\frac{10}{30}+\frac{9}{30}-\frac{15}{30}=\frac{6+10+9-15}{30}=\frac{10}{30}=\frac{1}{3}\)
2^x+2 *3^x+1*5^x=10800
2^x*2^2*3^x*2*5^x=10800
(2*3*5)^x*12=10800
30^x=10800