Đáp án:
Giải thích các bước giải:
$$\eqalign{
& a)\,\,\sqrt {{6 \over {x - 3}}} + \sqrt {{8 \over {2 - x}}} = 6 \cr
& DKXD:\,\,\left\{ \matrix{
x - 3 > 0 \hfill \cr
2 - x > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x > 3 \hfill \cr
x < 2 \hfill \cr} \right. \Rightarrow Vo\,\,nghiem \cr
& b)\,\,\sqrt {4x - 1} + \sqrt {4{x^2} - 1} = 1 \cr
& DKXD:\,\,\left\{ \matrix{
4x - 1 \ge 0 \hfill \cr
4{x^2} - 1 \ge 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x > {1 \over 4} \hfill \cr
\left[ \matrix{
x \ge {1 \over 2} \hfill \cr
x \le - {1 \over 2} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \ge {1 \over 2} \cr
& PT \Leftrightarrow 4x - 1 + 4{x^2} - 1 + 2\sqrt {\left( {4x - 1} \right)\left( {4{x^2} - 1} \right)} = 1 \cr
& \Leftrightarrow 4{x^2} + 4x + 1 + 2\sqrt {\left( {4x - 1} \right)\left( {4{x^2} - 1} \right)} = 4 \cr
& \Leftrightarrow {\left( {2x + 1} \right)^2} + 2\sqrt {\left( {4x - 1} \right)\left( {4{x^2} - 1} \right)} = 4 \cr
& x \ge {1 \over 2} \Leftrightarrow 2x \ge 1 \Leftrightarrow 2x + 1 \ge 2 \Leftrightarrow {\left( {2x + 1} \right)^2} \ge 4 \cr
& \Rightarrow VT \ge 4 = VP \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x = {1 \over 2} \cr
& Thu\,\,lai:\,\,\sqrt {4.{1 \over 2} - 1} + \sqrt {4.{1 \over 4} - 1} = 1\,\,\left( {dung} \right) \cr
& Vay\,\,x = {1 \over 2} \cr} $$
