Đáp án:
$\begin{array}{l}
1)H = \left| {x - 3} \right| + 2020\\
Do:\left| {x - 3} \right| \ge 0\forall x\\
\Rightarrow \left| {x - 3} \right| + 2020 \ge 2020\forall x\\
\Rightarrow H \ge 2020\forall x\\
\Rightarrow GTNN:H = 2020\\
Khi:x = 3\\
B = {\left( {x - 1} \right)^2} + 2021\\
Do:{\left( {x - 1} \right)^2} \ge 0\forall x\\
\Rightarrow {\left( {x - 1} \right)^2} + 2021 \ge 2021\forall x\\
\Rightarrow B \ge 2021\forall x\\
\Rightarrow GTNN:B = 2021\\
Khi:x = 1\\
C2)\\
\dfrac{{3{a^2} - {b^2}}}{{{a^2} + {b^2}}} = \dfrac{3}{4}\\
\Rightarrow 4.\left( {3{a^2} - {b^2}} \right) = 3.\left( {{a^2} + {b^2}} \right)\\
\Rightarrow 12{a^2} - 4{b^2} = 3{a^2} + 3{b^2}\\
\Rightarrow 12{a^2} - 3{a^2} = 3{b^2} + 4{b^2}\\
\Rightarrow 9{a^2} = 7{b^2}\\
\Rightarrow \dfrac{{{a^2}}}{{{b^2}}} = \dfrac{7}{9}\\
\text{Vậy}\,\dfrac{{{a^2}}}{{{b^2}}} = \dfrac{7}{9}\\
C3)\\
Thay\,ab = {c^2}\\
\Rightarrow \dfrac{{{a^2} + {c^2}}}{{{b^2} + {c^2}}} = \dfrac{{{a^2} + a.b}}{{{b^2} + a.b}} = \dfrac{{a.\left( {a + b} \right)}}{{b.\left( {a + b} \right)}} = \dfrac{a}{b}\\
\text{Vậy}\,\dfrac{{{a^2} + {c^2}}}{{{b^2} + {c^2}}} = \dfrac{a}{b}\\
C4)\\
\dfrac{a}{2} = \dfrac{b}{5} = \dfrac{c}{7} = k \Rightarrow \left\{ \begin{array}{l}
a = 2k\\
b = 5k\\
c = 7k
\end{array} \right.\\
A = \dfrac{{a - b + c}}{{a + 2b - c}} = \dfrac{{2k - 5k + 7k}}{{2k + 2.5k - 7k}}\\
= \dfrac{{4k}}{{5k}} = \dfrac{4}{5}\\
\text{Vậy}\,A = \dfrac{4}{5}
\end{array}$
