Đáp án:

`a)`
${\log ^2}_3x + 2{\log _3}x - 3 = 0$
$ \Leftrightarrow {\log ^2}_3x + 2{\log _3}x + 1 = 4$
$ \Leftrightarrow {({\log _3}x + 1)^2} = 4$
 $\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_3}x + 1 = 2}\\
{{{\log }_3}x + 1 =  - 2}
\end{array}} \right.\\$
$ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_3}x = 1}\\
{{{\log }_3}x =  - 3}
\end{array}} \right.\\$
 $\Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{{{\log }_3}x = {{\log }_3}3}\\
{{{\log }_3}x = {{\log }_3}\frac{1}{{27}}}
\end{array}} \right.\\$
$ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = 3}\\
{x = \frac{1}{{27}}}
\end{array}} \right.$
Vậy `x in{3; 1/27}` $\\$
`b)`
ĐKXĐ: `x>1`
$4{\log _9}x + {\log _x}3 = 3$
 $\Leftrightarrow 4{\log _9}\left( x \right) + \frac{{{{\log }_9}\left( 3 \right)}}{{{{\log }_9}\left( x \right)}} = 3$
 $\Leftrightarrow 4{\log _9}\left( x \right) + \frac{1}{{2{{\log }_9}\left( x \right)}} = 3$

Đặt `u=log_9 x`, phương trình trở thành: $4u+\frac{1}{2u}=3$

$ \Leftrightarrow 8{u^2} + 1 = 6u$
$\Leftrightarrow 8{u^2} - 6u + 1 = 0$

Ta có $Δ=\left(-6\right)^2-4\cdot \:8\cdot \:1=4>1$

$⇒ u_1=\frac{-\left(-6\right)\sqrt+4}{2\cdot \:8},\:u_2=\frac{-\left(-6\right)-\sqrt4}{2\cdot \:8}$

$⇒ u=\frac{1}{2}; u=\frac{1}{4}$

$⇒ \log_9 x=\frac{1}{2}; \log_9 x=\frac{1}{4}$

$⇒ x=\sqrt{9}; x=\sqrt3$

$⇒ x=3; x=\sqrt3$ (cùng thoả mãn)

Vậy `x\in{3; sqrt3}` $\\$

`c)`

ĐKXĐ: `x > 0`

${\log _2}^2x + 2{\log _2}\sqrt x  - 2 = 0  $
$\Leftrightarrow {\log _2}^2x + 2 \cdot \frac{1}{2}{\log _2}x - 2 = 0 $
$\Leftrightarrow {\log _2}^2x + {\log _2}x - 2 = 0$

Đặt `u=tan_2x`, phương trình trở thành `u^2+u-2=0`

Ta có: $Δ=1^2-4\cdot \:1\left(-2\right)=9>1$

$⇒ u_1=\frac{-1+ \sqrt{9}}{2\cdot \:1}; u_2=\frac{-1- \sqrt{9}}{2\cdot \:1}$

$⇒ u=1; u=-2$

$⇒ log_2x=1; log_2x=-2$

$⇒x=2; x=\frac{1}{4}$ (cùng thoả mãn)

Vậy `x\in{1/4; 2}`