Đáp án:
\(\begin{array}{l}
B19:\\
a)x = \dfrac{7}{2}\\
b)x = \dfrac{3}{2}\\
B20:\\
a)x = 3\\
b)x = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B19:\\
a)\sqrt {{x^2} - 6x + 9} = 4 - x\\
\to \sqrt {{{\left( {x - 3} \right)}^2}} = 4 - x\\
\to \left| {x - 3} \right| = 4 - x\\
\to \left[ \begin{array}{l}
x - 3 = 4 - x\\
x - 3 = - 4 + x
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 7\\
- 3 = - 4\left( l \right)
\end{array} \right.\\
\to x = \dfrac{7}{2}\\
b)DK:x \ge \dfrac{3}{2}\\
\sqrt {2x - 3 + 2\sqrt {2x - 3} .1 + 1} + \sqrt {2x - 3 + 2\sqrt {2x - 3} .4 + 16} = 5\\
\to \sqrt {{{\left( {\sqrt {2x - 3} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {2x - 3} + 4} \right)}^2}} = 5\\
\to \left| {\sqrt {2x - 3} + 1} \right| + \left| {\sqrt {2x - 3} + 4} \right| = 5\\
\to \sqrt {2x - 3} + 1 + \sqrt {2x - 3} + 4 = 5\\
\to 2\sqrt {2x - 3} + 5 = 5\\
\to \sqrt {2x - 3} = 0\\
\to x = \dfrac{3}{2}\\
B20:\\
a)DK:x \ge 3\\
\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} + \sqrt {{{\left( {x - 3} \right)}^2}} = 0\\
\to \sqrt {x - 3} \left( {\sqrt {x + 3} + \sqrt {x - 3} } \right) = 0\\
\to \sqrt {x - 3} = 0\left( {do:\sqrt {x + 3} + \sqrt {x - 3} > 0\forall x \ge 3} \right)\\
\to x = 3\\
b)\sqrt {{{\left( {x - 1} \right)}^2}} + \sqrt {{{\left( {x - 2} \right)}^2}} = 3\\
\to \left| {x - 1} \right| + \left| {x - 2} \right| = 3\\
\to \left[ \begin{array}{l}
x - 1 + x - 2 = 3\left( {DK:x \ge 2} \right)\\
x - 1 - x + 2 = 3\left( {DK:2 > x \ge 1} \right)\\
- x + 1 - x + 2 = 3\left( {DK:1 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 6\\
1 = 3\left( l \right)\\
- 2x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
