Đáp án:
\(\begin{array}{l}
b,\\
B = - 2\\
c,\\
C = - 1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
b,\\
x > \dfrac{1}{2} \Leftrightarrow 2x > 1 \Leftrightarrow 2x - 1 > 0 \Rightarrow \left| {2x - 1} \right| = 2x - 1\\
B = 2x - 3 - \sqrt {4{x^2} - 4x + 1} \\
= 2x - 3 - \sqrt {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \\
= 2x - 3 - \sqrt {{{\left( {2x - 1} \right)}^2}} \\
= 2x - 3 - \left| {2x - 1} \right|\\
= 2x - 3 - \left( {2x - 1} \right)\\
= 2x - 3 - 2x + 1\\
= - 2\\
c,\\
x < 5 \Rightarrow x - 5 < 0 \Rightarrow \left| {x - 5} \right| = - \left( {x - 5} \right)\\
C = \dfrac{{\sqrt {{x^2} - 10x + 25} }}{{x - 5}}\\
= \dfrac{{\sqrt {{x^2} - 2.x.5 + {5^2}} }}{{x - 5}}\\
= \dfrac{{\sqrt {{{\left( {x - 5} \right)}^2}} }}{{x - 5}}\\
= \dfrac{{\left| {x - 5} \right|}}{{x - 5}}\\
= \dfrac{{ - \left( {x - 5} \right)}}{{x - 5}}\\
= - 1
\end{array}\)
