Đáp án:
Bài 1:
$\begin{array}{l}
12)\left\{ \begin{array}{l}
9x,khi:0 \le x < 1\\
x,khi:x < 0
\end{array} \right.\\
13) - 1\\
14)\dfrac{1}{y}\\
15) - x + 1
\end{array}$
Bài 3: $A=4042$
Giải thích các bước giải:
$\begin{array}{l}
\text{Bài 1}:\\
12)5x + 4\sqrt {{x^2}} \\
= 5x + 4\left| x \right|\\
= \left\{ \begin{array}{l}
5x + 4x,khi:0 \le x < 1\\
5x - 4x,khi:x < 0
\end{array} \right.\\
= \left\{ \begin{array}{l}
9x,khi:0 \le x < 1\\
x,khi:x < 0
\end{array} \right.\\
13) - \dfrac{{\sqrt {{x^2}} }}{x} = - \dfrac{{\left| x \right|}}{x} = - \dfrac{x}{x} = - 1\left( {do:x > 0 \Rightarrow \left| x \right| = x} \right)\\
14)\dfrac{y}{x}\sqrt {\dfrac{{{x^2}}}{{{y^4}}}} = \dfrac{y}{x}\left| {\dfrac{x}{{{y^2}}}} \right| = \dfrac{y}{x}.\dfrac{x}{{{y^2}}} = \dfrac{1}{y}\\
15)\left| {x - 2} \right| + \dfrac{{\sqrt {{x^2} - 4x + 4} }}{{x - 2}}\\
= \left| {x - 2} \right| + \dfrac{{\sqrt {{{\left( {x - 2} \right)}^2}} }}{{x - 2}}\\
= \left| {x - 2} \right| + \dfrac{{\left| {x - 2} \right|}}{{x - 2}}\\
= - \left( {x - 2} \right) + \dfrac{{ - \left( {x - 2} \right)}}{{x - 2}}\left( {do:x < 2} \right)\\
= - x + 2 - 1\\
= - x + 1
\end{array}$
$\text{Bài 3:}$
Ta có:
$A = x\sqrt {\dfrac{{\left( {2021 + {y^2}} \right)\left( {2021 + {z^2}} \right)}}{{\left( {2021 + {x^2}} \right)}}} + y\sqrt {\dfrac{{\left( {2021 + {z^2}} \right)\left( {2021 + {x^2}} \right)}}{{2021 + {y^2}}}} + z\sqrt {\dfrac{{\left( {2021 + {x^2}} \right)\left( {2021 + {y^2}} \right)}}{{2021 + {z^2}}}} $
Nhận xét:
$\begin{array}{l}
x\sqrt {\dfrac{{\left( {2021 + {y^2}} \right)\left( {2021 + {z^2}} \right)}}{{\left( {2021 + {x^2}} \right)}}} \\
= x\sqrt {\dfrac{{\left( {xy + yz + zx + {y^2}} \right)\left( {xy + yz + zx + {z^2}} \right)}}{{\left( {xy + yz + zx + {x^2}} \right)}}} \\
= x\sqrt {\dfrac{{\left( {y + x} \right)\left( {y + z} \right)\left( {z + y} \right)\left( {z + x} \right)}}{{\left( {x + y} \right)\left( {x + z} \right)}}} \\
= x\sqrt {{{\left( {y + z} \right)}^2}} \\
= x\left| {y + z} \right|\\
= x\left( {y + z} \right)\left( {do:y,z > 0} \right)
\end{array}$
Hoàn toàn tương tự ta có:
$\left\{ \begin{array}{l}
y\sqrt {\dfrac{{\left( {2021 + {z^2}} \right)\left( {2021 + {x^2}} \right)}}{{2021 + {y^2}}}} = y\left( {x + z} \right)\\
z\sqrt {\dfrac{{\left( {2021 + {x^2}} \right)\left( {2021 + {y^2}} \right)}}{{2021 + {z^2}}}} = z\left( {x + y} \right)
\end{array} \right.$
Khi đó:
$\begin{array}{l}
A = x\left( {y + z} \right) + y\left( {x + z} \right) + z\left( {x + y} \right)\\
= 2\left( {xy + yz + xz} \right)\\
= 2.2021\\
= 4042
\end{array}$
