a) \(\displaystyle {lo{g_3}\left( {5x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}lo{g_3}\left( {7x{\rm{ }} + {\rm{ }}5} \right)}\) (1)
TXĐ: \(\displaystyle D = \left( {{{ - 3} \over 5}, + \infty } \right)\)
Khi đó: (1) \(\displaystyle ⇔ 5x + 3 = 7x + 5 \) \(\displaystyle ⇔2x=-2 ⇔ x = -1\) (loại)
Vậy phương trình (1) vô nghiệm.
b) \(\displaystyle {\log\left( {x{\rm{ }}-{\rm{ }}1} \right){\rm{ }}-{\rm{ }}\log\left( {2x{\rm{ }}-{\rm{ }}11} \right){\rm{ }} = {\rm{ }}\log{\rm{ }}2}\) (2)
TXĐ: \(\displaystyle D = \left( {\dfrac{{11}}{2}; + \infty } \right).\)
Khi đó: \(\displaystyle (2) \Leftrightarrow \log {{x - 1} \over {2x - 11}} = \log 2\) \(\displaystyle \Leftrightarrow {{x - 1} \over {2x - 11}} = 2\) \(\displaystyle \Rightarrow x - 1 = 4x - 22 \Leftrightarrow 3x=21\) \(\displaystyle \Leftrightarrow x = 7 (TM)\)
Vậy phương trình có nghiệm là \(\displaystyle x = 7.\)
c) \(\displaystyle {lo{g_2}\left( {x{\rm{ }}-{\rm{ }}5} \right){\rm{ }} + {\rm{ }}lo{g_2}\left( {x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }}3}\) (3)
TXĐ: \(\displaystyle (5, +∞)\)
Khi đó:
\(\displaystyle (3) \, \Leftrightarrow {\log _2}(x - 5)(x + 2)=3\)
\(\displaystyle \Leftrightarrow \left( {x - 5} \right)(x + 2) = 2^3 \)
\(\displaystyle \Leftrightarrow {x^2} - 3x - 18 = 0 \\ \Leftrightarrow (x-6)(x+3)=0 \\ \Leftrightarrow \left[ \matrix{
x - 6=0 \hfill \cr
x + 3=0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 6 \, \, (tm) \hfill \cr
x = - 3 \, \,(ktm) \hfill \cr} \right.\)
Vậy phương trình có nghiệm \(\displaystyle x = 6\)
d) \(\displaystyle {log{\rm{ }}\left( {{x^2}-{\rm{ }}6x{\rm{ }} + {\rm{ }}7} \right){\rm{ }} = {\rm{ }}log{\rm{ }}\left( {x{\rm{ }}-{\rm{ }}3} \right)}\) (4)
TXĐ: \(\displaystyle D = (3 + \sqrt 2 , + \infty )\)
Khi đó:
\(\displaystyle \begin{array}{l}\left( 4 \right) \Leftrightarrow {x^2} - 6x + 7 = x - 3\\ \Leftrightarrow {x^2} - 7x + 10 = 0\\ \Leftrightarrow \left( {x - 5} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 5 = 0\\x - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 5\;\;\left( {tm} \right)\\x = 2\;\;\left( {ktm} \right)\end{array} \right..\end{array}\)
Vậy phương trình có nghiệm là \(\displaystyle x = 5\).
