`#3107.101107`
`a)`
`8/25 = (2^x)/(5^(x - 1))`
`(2^3)/(5^2) = (2^x)/(5^(x - 1))`
\(\left\{{}\begin{matrix}x=3\\x-1=2\end{matrix}\right.\\ \left\{{}\begin{matrix}x=3\\x=2+1=3\end{matrix}\right.\)
Vậy, `x = 3`
`b)`
`x^2 - 3/4x = 0`
`x(x - 3/4) = 0`
TH1: `x = 0`
TH2: `x - 3/4 = 0`
`x = 3/4`
Vậy, `x \in {0; 3/4}.`
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