Câu 1: $\log5+\log(x+10)-1=\log(21x-20)-\log(2x-1)$
Câu 2: $\ln(x^2-6x+7)=\ln(x-3)$
Câu 1:
ĐK: $\begin{cases} x+10>0\\21x-20>0\\2x-1>0\end{cases}\Leftrightarrow x>\dfrac{20}{21}$
$\log5+\log(x+10)-1=\log(21x-20)-\log(2x-1)$
$\Leftrightarrow\log5+\log(x+10)+\log(2x-1)=\log(21x-20)+1$
$\Leftrightarrow\log[5(x+10)(2x-1)]=\log[10(21x-20)]$
$\Leftrightarrow5(x+10)(2x-1)=10(21x-20)$
$\Leftrightarrow2x^2-23x+30=0$
$\Leftrightarrow\left[\begin{array}{l}x=\dfrac{3}2 &(n)\\x=10&(n)\end{array}\right.$
Câu 2:
ĐK: $\begin{cases}x^2-6x+7>0\\x-3>0\end{cases}\Leftrightarrow x>3+\sqrt2$
$\ln(x^2-6x+7)=\ln(x-3)$
$\Leftrightarrow x^2-6x+7=x-3$
$\Leftrightarrow x^2-7x+10=0$
$\Leftrightarrow\left[\begin{array}{l}x=5&(n)\\x=2&(l)\end{array}\right.$
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