Đáp án:
i. \(\left[ \begin{array}{l}
B = 4x - 4\\
B = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Do:x < 0\\
\sqrt {9{x^2}} - 2x = 3\left| x \right| - 2x\\
= 3\left( { - x} \right) - 2x = - 5x\\
c.Do:x < 2\\
3\sqrt {{{\left( {x - 2} \right)}^2}} = 3\left| {x - 2} \right|\\
= 3\left[ { - \left( {x - 2} \right)} \right]\\
= - 3\left( {x - 2} \right) = - 3x + 6\\
e.Do:x \ge 0\\
5\left| x \right| + 3x = 5x + 3x = 8x\\
g.x - 4 + \sqrt {{{\left( {4 - x} \right)}^2}} = x - 4 + \left| {4 - x} \right|\\
= x - 4 - \left( {4 - x} \right)\left( {do:x > 4} \right)\\
= x - 4 - 4 + x\\
= 2x - 8\\
h.A = \sqrt {{{\left( {1 - 2a} \right)}^2}} - 2a\\
= \left| {1 - 2a} \right| - 2a\\
\to \left[ \begin{array}{l}
A = 1 - 2a - 2a\left( {DK:\dfrac{1}{2} \ge a} \right)\\
A = - 1 + 2a - 2a\left( {DK:\dfrac{1}{2} < a} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = 1 - 4a\\
A = - 1
\end{array} \right.\\
b.2\sqrt {{x^2}} = 2\left| x \right| = 2x\left( {do:x \ge 0} \right)\\
d.2\sqrt {{x^2}} - 5x = 2\left| x \right| - 5x\left( {do:x < 0} \right)\\
= - 2x - 5x = - 7x\\
f.3{x^2} + 3{x^2} = 6{x^2}\\
i.B = \sqrt {{{\left( {2x - 3} \right)}^2}} + 2x - 1\\
= \left| {2x - 3} \right| + 2x - 1\\
\to \left[ \begin{array}{l}
B = 2x - 3 + 2x - 1\left( {DK:x \ge \dfrac{3}{2}} \right)\\
B = - 2x + 3 + 2x - 1\left( {DK:x < \dfrac{3}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
B = 4x - 4\\
B = 2
\end{array} \right.
\end{array}\)
