Đáp án:
\(\begin{array}{l}
b)x = 1\\
c)x = \dfrac{9}{4}\\
d)x = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{b)\sqrt {{{\left( {x - 4} \right)}^2}} {\rm{ \;}} = x + 2}\\
{ \to \left| {x - 4} \right| = x + 2}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x - 4 = x + 2\left( {DK:x \ge 4} \right)}\\
{x - 4 = - x - 2\left( {DK:x < 4} \right)}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{ - 4 = 2\left( l \right)}\\
{2x = 2}
\end{array}} \right.}\\
{ \to x = 1}\\
{c)\sqrt {{{\left( {x - 3} \right)}^2}} {\rm{ \;}} = 3x - 6}\\
{ \to \left| {x - 3} \right| = 3x - 6}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x - 3 = 3x - 6\left( {DK:x \ge 3} \right)}\\
{x - 3 = {\rm{ \;}} - 3x + 6\left( {DK:x < 3} \right)}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{2x = 3}\\
{4x = 9}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = \dfrac{3}{2}\left( l \right)}\\
{x = \dfrac{9}{4}}
\end{array}} \right.}\\
{ \to x = \dfrac{9}{4}}\\
{d)\sqrt {{{\left( {x - 2} \right)}^2}} {\rm{ \;}} = 2x - 5}\\
{ \to \left| {x - 2} \right| = 2x - 5}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x - 2 = 2x - 5\left( {DK:x \ge 2} \right)}\\
{x - 2 = {\rm{ \;}} - 2x + 5\left( {DK:x < 2} \right)}
\end{array}} \right.}\\
{{\rm{\;}} \to \left[ {\begin{array}{*{20}{l}}
{x = 3}\\
{3x = 7}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 3}\\
{x = \dfrac{7}{3}\left( l \right)}
\end{array}} \right.}\\
{ \to x = 3}
\end{array}\)
