Đáp án:
$\begin{array}{l}
3)a)\\
x < \dfrac{1}{2}\\
\Rightarrow M = \left| {x - 1} \right| - \left| {1 - 2x} \right|\\
= 1 - x - \left( {1 - 2x} \right)\\
= x\\
b)x > \dfrac{1}{2}\\
\Rightarrow N = 2x - \sqrt {4{x^2} - 4x + 1} \\
= 2x - \sqrt {{{\left( {2x - 1} \right)}^2}} \\
= 2x - \left| {2x - 1} \right|\\
= 2x - \left( {2x - 1} \right)\\
= 1\\
c)P = \sqrt {{x^2} + 4x + 4} + \sqrt {{x^2}} \left( {x \ge 0} \right)\\
= \sqrt {{{\left( {x + 2} \right)}^2}} + \left| x \right|\\
= \left| {x + 2} \right| + x\\
= x + 2 + x\\
= 2x + 2\\
d)Q = \sqrt {x + 2\sqrt {x - 1} } - \sqrt {x - 1} + 4\left( {x \ge 1} \right)\\
= \sqrt {x - 1 + 2\sqrt {x - 1} + 1} - \sqrt {x - 1} + 4\\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} - \sqrt {x - 1} + 4\\
= \sqrt {x - 1} + 1 - \sqrt {x - 1} + 4\\
= 5\\
4a)x > 2\\
A = \left| {x + 2} \right| + \dfrac{{\sqrt {{x^2} - 4x + 4} }}{{x - 2}}\\
= x + 2 + \dfrac{{\sqrt {{{\left( {x - 2} \right)}^2}} }}{{x - 2}}\\
= x + 2 + \dfrac{{\left| {x - 2} \right|}}{{x - 2}}\\
= x + 2 + \dfrac{{x - 2}}{{x - 2}}\\
= x + 2 + 1\\
= x + 3\\
b)B = \left| {4 - x} \right| + \dfrac{{4 - x}}{{\sqrt {{x^2} - 8x + 16} }}\left( {x < 4} \right)\\
= 4 - x + \dfrac{{4 - x}}{{\sqrt {{{\left( {x - 4} \right)}^2}} }}\\
= 4 - x + \dfrac{{4 - x}}{{\left| {x - 4} \right|}}\\
= 4 - x + \dfrac{{4 - x}}{{4 - x}}\\
= 4 - x + 1\\
= 5 - x\\
c)C = \left| {x - 3} \right| - \dfrac{{3 - x}}{{\sqrt {9 - 6x + {x^2}} }}\\
= 3 - x - \dfrac{{3 - x}}{{\left| {x - 3} \right|}}\\
= 3 - x - \dfrac{{3 - x}}{{3 - x}}\\
= 3 - x - 1\\
= 2 - x
\end{array}$
