Toán Cho log (12) 18= a + b/(c+log (2) 3) .Tính tông a+b+c 05/12/2021 By Autumn Cho log (12) 18= a + b/(c+log (2) 3) .Tính tông a+b+c
$log_{12}18=\dfrac{log_218}{log_212}=\dfrac{log_2(2.3^2)}{log2(2^2.3)}=\dfrac{1+2log_23}{2+log_23}=\dfrac{4+2log_23-3}{2+log_23}=2+\dfrac{-3}{2+log_23}\\=a+\dfrac{b}{c+log_23}=>a+b+c=2+2-3=1$ Trả lời
`log_{12} 18 = a + b/(c + log_{2} 3)` `text{Ta có}` `log_{12} 18` `= (log_{2} 18)/(log_{2} 12)` `= (log_{2} (2.3^{2)))/(log_{2} (3.2^{2}))` `= (log_{2} 2 + 2log_{2} 3)/(log_{2} 2^{2} + log_{2} 3)` `= (1 + 2log_{2} 3)/(2 + log_{2} 3)` `= (4 + 2log_{2} 3 – 3)/(2 + log_{2} 3)` `= (2(2 + log_{2} 3))/(2 + log_{2} 3) + (-3)/(2 + log_{2} 3)` `= 2 + (-3)/(2 + log_{2} 3) = a + b/(c + log_{2} 3)` `->` \(\left\{ \begin{array}{l}a = 2\\b = -3\\c = 2\end{array} \right.\) `-> a + b + c = 2 – 3 + 2 = 1` Trả lời
$log_{12}18=\dfrac{log_218}{log_212}=\dfrac{log_2(2.3^2)}{log2(2^2.3)}=\dfrac{1+2log_23}{2+log_23}=\dfrac{4+2log_23-3}{2+log_23}=2+\dfrac{-3}{2+log_23}\\=a+\dfrac{b}{c+log_23}=>a+b+c=2+2-3=1$
`log_{12} 18 = a + b/(c + log_{2} 3)`
`text{Ta có}`
`log_{12} 18`
`= (log_{2} 18)/(log_{2} 12)`
`= (log_{2} (2.3^{2)))/(log_{2} (3.2^{2}))`
`= (log_{2} 2 + 2log_{2} 3)/(log_{2} 2^{2} + log_{2} 3)`
`= (1 + 2log_{2} 3)/(2 + log_{2} 3)`
`= (4 + 2log_{2} 3 – 3)/(2 + log_{2} 3)`
`= (2(2 + log_{2} 3))/(2 + log_{2} 3) + (-3)/(2 + log_{2} 3)`
`= 2 + (-3)/(2 + log_{2} 3) = a + b/(c + log_{2} 3)`
`->` \(\left\{ \begin{array}{l}a = 2\\b = -3\\c = 2\end{array} \right.\)
`-> a + b + c = 2 – 3 + 2 = 1`