Đáp án:
\(\begin{array}{l}
B2:\\
b)\left[ \begin{array}{l}
B = 2\sqrt {x - 2} \\
B = 2
\end{array} \right.\\
B3:\\
a)x = 2\\
c)x = 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
b)DK:x \ge 2\\
B = \sqrt {x - 2 + 2\sqrt {x - 2} .1 + 1} + \sqrt {x - 2 - 2\sqrt {x - 2} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {x - 2} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 2} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 2} + 1} \right| + \left| {\sqrt {x - 2} - 1} \right|\\
= \sqrt {x - 2} + 1 + \left| {\sqrt {x - 2} - 1} \right|\\
\to \left[ \begin{array}{l}
B = \sqrt {x - 2} + 1 + \sqrt {x - 2} - 1\left( {DK:x \ge 3} \right)\\
B = \sqrt {x - 2} + 1 - \sqrt {x - 2} + 1\left( {DK:2 \le x < 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
B = 2\sqrt {x - 2} \\
B = 2
\end{array} \right.\\
B3:\\
a)\sqrt {{x^2} - 2x + 4} = 2x - 2\\
\to {x^2} - 2x + 4 = 4{x^2} - 8x + 4\left( {DK:x \ge 1} \right)\\
\to 3{x^2} - 6x = 0\\
\to 3x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x = 2
\end{array} \right.\\
\to x = 2\\
c)DK:x \ge 3\\
\sqrt {x - 3} - 2\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} = 0\\
\to \sqrt {x - 3} \left( {1 - 2\sqrt {x + 3} } \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
1 - 2\sqrt {x + 3} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
2\sqrt {x + 3} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
4\left( {x + 3} \right) = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{{11}}{4}\left( l \right)
\end{array} \right.
\end{array}\)
