Đáp án:

\(\begin{array}{l}
1,\\
a,\\
\left[ \begin{array}{l}
x = \dfrac{7}{4}\\
x =  - \dfrac{7}{4}
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x =  - 3\\
x = \dfrac{1}{3}
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = 10\\
x =  - 2
\end{array} \right.\\
d,\\
x =  - \dfrac{{65}}{{12}}\\
e,\\
x =  - \dfrac{4}{3}\\
2,\\
a,\\
\left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.\\
b,\\
\left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = 3
\end{array} \right.\\
c,\\
\left\{ \begin{array}{l}
x = 5\\
y =  - 10
\end{array} \right.\\
d,\\
\left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\\
3,\\
a,\\
A = 16\\
b,\\
B = 145\\
4,\\
a,\\
C = 3\\
b,\\
D = 69
\end{array}\) 

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
16{x^2} - 49 = 0\\
 \Leftrightarrow 16{x^2} = 49\\
 \Leftrightarrow {x^2} = \dfrac{{49}}{{16}}\\
 \Leftrightarrow {x^2} = \dfrac{{{7^2}}}{{{4^2}}}\\
 \Leftrightarrow {x^2} = {\left( {\dfrac{7}{4}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{4}\\
x =  - \dfrac{7}{4}
\end{array} \right.\\
b,\\
{\left( {x - 2} \right)^2} = 4{x^2} + 4x + 1\\
 \Leftrightarrow {\left( {x - 2} \right)^2} = {\left( {2x} \right)^2} + 2.2x.1 + {1^2}\\
 \Leftrightarrow {\left( {x - 2} \right)^2} = {\left( {2x + 1} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 2x + 1\\
x - 2 =  - \left( {2x + 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 - 2x - 1 = 0\\
x - 2 + 2x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - x - 3 = 0\\
3x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - x = 3\\
3x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = \dfrac{1}{3}
\end{array} \right.\\
c,\\
{x^2} - 8x = 20\\
 \Leftrightarrow {x^2} - 8x + 16 = 20 + 16\\
 \Leftrightarrow {x^2} - 2.x.4 + {4^2} = 36\\
 \Leftrightarrow {\left( {x - 4} \right)^2} = {6^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 6\\
x - 4 =  - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6 + 4\\
x =  - 6 + 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 10\\
x =  - 2
\end{array} \right.\\
d,\\
{\left( {2x + 1} \right)^2} - 4x.\left( {x - 2} \right) + 64 = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}} \right] - \left( {4{x^2} - 8x} \right) + 64 = 0\\
 \Leftrightarrow \left( {4{x^2} + 4x + 1} \right) - \left( {4{x^2} - 8x} \right) + 64 = 0\\
 \Leftrightarrow 4{x^2} + 4x + 1 - 4{x^2} + 8x + 64 = 0\\
 \Leftrightarrow \left( {4{x^2} - 4{x^2}} \right) + \left( {4x + 8x} \right) + \left( {1 + 64} \right) = 0\\
 \Leftrightarrow 12x + 65 = 0\\
 \Leftrightarrow x =  - \dfrac{{65}}{{12}}\\
e,\\
\left( {x - 3} \right)\left( {x + 3} \right) - {\left( {x + 2} \right)^2} = 2x - 5\\
 \Leftrightarrow \left( {{x^2} - {3^2}} \right) - \left( {{x^2} + 2.x.2 + {2^2}} \right) - \left( {2x - 5} \right) = 0\\
 \Leftrightarrow \left( {{x^2} - 9} \right) - \left( {{x^2} + 4x + 4} \right) - 2x + 5 = 0\\
 \Leftrightarrow {x^2} - 9 - {x^2} - 4x - 4 - 2x + 5 = 0\\
 \Leftrightarrow \left( {{x^2} - {x^2}} \right) + \left( { - 4x - 2x} \right) + \left( { - 9 - 4 + 5} \right) = 0\\
 \Leftrightarrow  - 6x - 8 = 0\\
 \Leftrightarrow  - 6x = 8\\
 \Leftrightarrow x =  - \dfrac{4}{3}\\
2,\\
a,\\
{x^2} - 6x + {y^2} - 4y + 13 = 0\\
 \Leftrightarrow \left( {{x^2} - 6x + 9} \right) + \left( {{y^2} - 4y + 4} \right) = 0\\
 \Leftrightarrow \left( {{x^2} - 2.x.3 + {3^2}} \right) + \left( {{y^2} - 2.y.2 + {2^2}} \right) = 0\\
 \Leftrightarrow {\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {x - 3} \right)^2} \ge 0,\,\,\forall x\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {x - 3} \right)^2} + {\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {x - 3} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 3 = 0\\
y - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.\\
b,\\
4{x^2} + 2{y^2} - 4xy = 6y - 9\\
 \Leftrightarrow 4{x^2} + 2{y^2} - 4xy - 6y + 9 = 0\\
 \Leftrightarrow \left( {4{x^2} - 4xy + {y^2}} \right) + \left( {{y^2} - 6y + 9} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - 2.2x.y + {y^2}} \right] + \left( {{y^2} - 2.y.3 + {3^2}} \right) = 0\\
 \Leftrightarrow {\left( {2x - y} \right)^2} + {\left( {y - 3} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {2x - y} \right)^2} \ge 0,\,\,\forall x,y\\
{\left( {y - 3} \right)^2} \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {2x - y} \right)^2} + {\left( {y - 3} \right)^2} \ge 0,\,\,\forall x,y\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {2x - y} \right)^2} = 0\\
{\left( {y - 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - y = 0\\
y - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x = y\\
y = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = 3
\end{array} \right.\\
c,\\
5{x^2} + 4xy + {y^2} - 10x + 25 = 0\\
 \Leftrightarrow \left( {4{x^2} + 4xy + {y^2}} \right) + \left( {{x^2} - 10x + 25} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} + 2.2x.y + {y^2}} \right] + \left( {{x^2} - 2.x.5 + {5^2}} \right) = 0\\
 \Leftrightarrow {\left( {2x + y} \right)^2} + {\left( {x - 5} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {2x + y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x - 5} \right)^2} \ge 0,\,\,\,\forall x
\end{array} \right\} \Rightarrow {\left( {2x + y} \right)^2} + {\left( {x - 5} \right)^2} \ge 0,\,\,\forall x,y\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {2x + y} \right)^2} = 0\\
{\left( {x - 5} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x + y = 0\\
x - 5 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y =  - 2x\\
x = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
y =  - 10
\end{array} \right.\\
d,\\
2{x^2} + 10{y^2} - 6x - 2y - 6xy + 10 = 0\\
 \Leftrightarrow \left( {{x^2} - 6xy + 9{y^2}} \right) + \left( {{x^2} - 6x + 9} \right) + \left( {{y^2} - 2y + 1} \right) = 0\\
 \Leftrightarrow \left[ {{x^2} - 2.x.3y + {{\left( {3y} \right)}^2}} \right] + \left( {{x^2} - 2.x.3 + {3^2}} \right) + \left( {{y^2} - 2.y.1 + {1^2}} \right) = 0\\
 \Leftrightarrow {\left( {x - 3y} \right)^2} + {\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {x - 3y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\forall x\\
{\left( {y - 1} \right)^2} \ge 0,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {x - 3y} \right)^2} + {\left( {x - 3} \right)^2} + {\left( {y - 1} \right)^2} \ge 0,\,\,\forall x,y\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {x - 3y} \right)^2} = 0\\
{\left( {x - 3} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 3y = 0\\
x - 3 = 0\\
y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\\
3,\\
a,\\
A = 2{a^2} + 5a + 4ab + 5b + 2{b^2} - 59\\
 = \left( {2{a^2} + 4ab + 2{b^2}} \right) + \left( {5a + 5b} \right) - 59\\
 = 2.\left( {{a^2} + 2.a.b + {b^2}} \right) + 5.\left( {a + b} \right) - 59\\
 = 2.{\left( {a + b} \right)^2} + 5.\left( {a + b} \right) - 59\\
a + b = 5 \Rightarrow A = {2.5^2} + 5.5 - 59 = 2.25 + 25 - 59 = 16\\
b,\\
B = 6{a^2} - 10b - 10a + 12ab + 6{b^2} + 45\\
 = \left( {6{a^2} + 12ab + 6{b^2}} \right) + \left( { - 10b - 10a} \right) + 45\\
 = 6.\left( {{a^2} + 2ab + {b^2}} \right) - 10\left( {a + b} \right) + 45\\
 = 6.{\left( {a + b} \right)^2} - 10\left( {a + b} \right) + 45\\
a + b = 5 \Rightarrow B = {6.5^2} - 10.5 + 45 = 145\\
4,\\
a,\\
C =  - 3x + {x^2} + 9y - 6xy + 9{y^2} - 1\\
 = \left( { - 3x + 9y} \right) + \left( {{x^2} - 6xy + 9{y^2}} \right) - 1\\
 =  - 3.\left( {x - 3y} \right) + \left[ {{x^2} - 2.x.3y + {{\left( {3y} \right)}^2}} \right] - 1\\
 =  - 3.\left( {x - 3y} \right) + {\left( {x - 3y} \right)^2} - 1\\
x - 3y = 4 \Rightarrow C =  - 3.4 + {4^2} - 1 = 3\\
b,\\
D = 3{x^2} + 2x - 18xy - 6y + 27{y^2} + 13\\
 = \left( {3{x^2} - 18xy + 27{y^2}} \right) + \left( {2x - 6y} \right) + 13\\
 = 3.\left( {{x^2} - 6xy + 9{y^2}} \right) + 2.\left( {x - 3y} \right) + 13\\
 = 3.\left[ {{x^2} - 2.x.3y + {{\left( {3y} \right)}^2}} \right] + 2.\left( {x - 3y} \right) + 13\\
 = 3.{\left( {x - 3y} \right)^2} + 2.\left( {x - 3y} \right) + 13\\
x - 3y = 4 \Rightarrow D = {3.4^2} + 2.4 + 13 = 69
\end{array}\)