a) \({lo{g_3}\left( {5x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}lo{g_3}\left( {7x{\rm{ }} + {\rm{ }}5} \right)}\)

Phương pháp giải:

+) Tìm điều kiện xác định.

+) Đưa về cùng cơ số: \({\log _a}f\left( x \right) = {\log _a}g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l} f\left( x \right) > 0\\ g\left( x \right) > 0\\ f\left( x \right) = g\left( x \right)\end{array} \right.\)

Lời giải chi tiết:

\(\displaystyle {lo{g_3}\left( {5x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}lo{g_3}\left( {7x{\rm{ }} + {\rm{ }}5} \right)}\) (1)

\(DK:\left\{ \begin{array}{l}5x + 3 > 0\\7x + 5 > 0\end{array} \right.

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\Leftrightarrow \left\{ \begin{array}{l}x > - \dfrac{3}{5}\\x > - \dfrac{5}{7}\end{array} \right. \) \(\Leftrightarrow x > - \dfrac{3}{5}\)

TXĐ: \(\displaystyle D = \left( {{{ - 3} \over 5}, + \infty } \right)\)

Khi đó: (1) \(\displaystyle \Rightarrow 5x + 3 = 7x + 5 \) \(\displaystyle ⇔2x=-2 ⇔ x = -1\) (loại)

Vậy phương trình (1) vô nghiệm.


LG b

b) \({\log \left( {x{\rm{ }}-{\rm{ }}1} \right){\rm{ }}-{\rm{ }}\log \left( {2x{\rm{ }}-{\rm{ }}11} \right){\rm{ }} = {\rm{ }}\log {\rm{ }}2}\)

Lời giải chi tiết:

\(\displaystyle {\log\left( {x{\rm{ }}-{\rm{ }}1} \right){\rm{ }}-{\rm{ }}\log\left( {2x{\rm{ }}-{\rm{ }}11} \right){\rm{ }} = {\rm{ }}\log{\rm{ }}2}\) (2)

\(DK:\left\{ \begin{array}{l}x - 1 > 0\\2x - 11 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x > 1\\x > \dfrac{{11}}{2}\end{array} \right. \) \(\Leftrightarrow x > \dfrac{{11}}{2}\)

TXĐ: \(\displaystyle D = \left( {\dfrac{{11}}{2}; + \infty } \right).\)

Khi đó: \(\displaystyle (2) \Rightarrow \log {{x - 1} \over {2x - 11}} = \log 2\) \(\displaystyle \Leftrightarrow {{x - 1} \over {2x - 11}} = 2\) \(\displaystyle \Rightarrow x - 1 = 4x - 22 \Leftrightarrow 3x=21\) \(\displaystyle \Leftrightarrow x = 7 (TM)\)

Vậy phương trình có nghiệm là \(\displaystyle x = 7.\)


LG c

c) \({lo{g_2}\left( {x{\rm{ }}-{\rm{ }}5} \right){\rm{ }} + {\rm{ }}lo{g_2}\left( {x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }}3}\)

Lời giải chi tiết:

\(\displaystyle {lo{g_2}\left( {x{\rm{ }}-{\rm{ }}5} \right){\rm{ }} + {\rm{ }}lo{g_2}\left( {x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }}3}\) (3)

\(DK:\left\{ \begin{array}{l}x - 5 > 0\\x + 2 > 0\end{array} \right.

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\) \(\Leftrightarrow \left\{ \begin{array}{l}x > 5\\x > - 2\end{array} \right. \Leftrightarrow x > 5\)

TXĐ: \(\displaystyle (5; +∞)\)

Khi đó:

\(\displaystyle (3) \, \Leftrightarrow {\log _2}<(x - 5)(x + 2)>=3\)

\(\displaystyle \Leftrightarrow \left( {x - 5} \right)(x + 2) = 2^3 \)

\( \Leftrightarrow {x^2} - 3x - 10 = 8\)

\(\displaystyle \Leftrightarrow {x^2} - 3x - 18 = 0 \\ \Leftrightarrow (x-6)(x+3)=0 \\ \Leftrightarrow \left< \matrix{x - 6=0 \hfill \cr x + 3=0 \hfill \cr} \right. \Leftrightarrow \left< \matrix{x = 6 \, \, (tm) \hfill \cr x = - 3 \, \,(ktm) \hfill \cr} \right.\)

Vậy phương trình có nghiệm \(\displaystyle x = 6\)


LG d

d) \({\log {\rm{ }}\left( {{x^2}-{\rm{ }}6x{\rm{ }} + {\rm{ }}7} \right){\rm{ }} = {\rm{ }}\log {\rm{ }}\left( {x{\rm{ }}-{\rm{ }}3} \right)}\)

Lời giải chi tiết:

\(\displaystyle {log{\rm{ }}\left( {{x^2}-{\rm{ }}6x{\rm{ }} + {\rm{ }}7} \right){\rm{ }} = {\rm{ }}log{\rm{ }}\left( {x{\rm{ }}-{\rm{ }}3} \right)}\) (4)

\(DK:\left\{ \begin{array}{l}{x^2} - 6x + 7 > 0\\x - 3 > 0\end{array} \right. \) \(\Leftrightarrow \left\{ \begin{array}{l}\left< \begin{array}{l}x > 3 + \sqrt 2 \\x \end{array} \right.\\x > 3\end{array} \right.\) \( \Leftrightarrow x > 3 + \sqrt 2 \)

TXĐ: \(\displaystyle D = (3 + \sqrt 2 , + \infty )\)

Khi đó:

\(\displaystyle \begin{array}{l}\left( 4 \right) \Leftrightarrow {x^2} - 6x + 7 = x - 3\\\Leftrightarrow {x^2} - 7x + 10 = 0\\\Leftrightarrow \left( {x - 5} \right)\left( {x - 2} \right) = 0\\\Leftrightarrow \left< \begin{array}{l}x - 5 = 0\\x - 2 = 0\end{array} \right. \Leftrightarrow \left< \begin{array}{l}x = 5\;\;\left( {tm} \right)\\x = 2\;\;\left( {ktm} \right)\end{array} \right..\end{array}\)