Đáp án:

\(\begin{array}{l}
3,\\
{a^2} - {b^2} = 119\\
{a^4} + {b^4} = 21361\\
4,\\
x =  - y\\
5,\\
a = b = c\\
6,\\
3)\,\,\left\{ \begin{array}{l}
x = 2\\
y =  - 1
\end{array} \right.\\
4)\,\,\,Phương\,\,trình\,\,vô\,\,nghiệm\\
5)\,\,\,Phương\,\,trình\,\,vô\,\,nghiệm
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
3,\\
{\left( {a - b} \right)^2} + 4ab = {7^2} + 4.60\\
 \Leftrightarrow {a^2} - 2ab + {b^2} + 4ab = 289\\
 \Leftrightarrow {a^2} + 2ab + {b^2} = 289\\
 \Leftrightarrow {\left( {a + b} \right)^2} = {17^2}\\
a > b > 0 \Rightarrow a + b > 0\\
 \Rightarrow a + b = 17\\
{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right) = 7.17 = 119\\
{a^4} + {b^4} = \left( {{a^4} - 2{a^2}{b^2} + {b^4}} \right) + 2{a^2}{b^2}\\
 = \left[ {{{\left( {{a^2}} \right)}^2} - 2{a^2}.{b^2} + {{\left( {{b^2}} \right)}^2}} \right] + 2.{\left( {ab} \right)^2}\\
 = \left( {{a^2} - {b^2}} \right) + 2.{\left( {ab} \right)^2}\\
 = {119^2} + {2.60^2}\\
 = 21361\\
4,\\
2\left( {{x^2} + {y^2}} \right) = {\left( {x - y} \right)^2}\\
 \Leftrightarrow 2{x^2} + 2{y^2} = {x^2} - 2xy + {y^2}\\
 \Leftrightarrow 2{x^2} + 2{y^2} - {x^2} + 2xy - {y^2} = 0\\
 \Leftrightarrow {x^2} + 2xy + {y^2} = 0\\
 \Leftrightarrow {\left( {x + y} \right)^2} = 0\\
 \Leftrightarrow x + y = 0\\
 \Leftrightarrow x =  - y\\
5,\\
{a^2} + {b^2} + {c^2} = ab + bc + ca\\
 \Leftrightarrow 2.\left( {{a^2} + {b^2} + {c^2}} \right) = 2.\left( {ab + bc + ca} \right)\\
 \Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} = 2ab + 2bc + 2ca\\
 \Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0\\
 \Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca + {a^2}} \right) = 0\\
 \Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\,\,\,\,\,\left( 1 \right)\\
{\left( {a - b} \right)^2} \ge 0,\,\,\,\forall a,b\\
{\left( {b - c} \right)^2} \ge 0,\,\,\,\,\forall b,c\\
{\left( {c - a} \right)^2} \ge 0,\,\,\,\forall c,a\\
 \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0,\,\,\,\forall a,b,c\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {a - b} \right)^2} = 0\\
{\left( {b - c} \right)^2} = 0\\
{\left( {c - a} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a - b = 0\\
b - c = 0\\
c - a = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = b\\
b = c\\
c = a
\end{array} \right. \Leftrightarrow a = b = c\\
6,\\
3)\,\,{x^2} + 3{y^2} - 4x + 6y + 7 = 0\\
 \Leftrightarrow \left( {{x^2} - 4x + 4} \right) + \left( {3{y^2} + 6y + 3} \right) = 0\\
 \Leftrightarrow \left( {{x^2} - 4x + 4} \right) + 3.\left( {{y^2} + 2y + 1} \right) = 0\\
 \Leftrightarrow \left( {{x^2} - 2.x.2 + {2^2}} \right) + 3.\left( {{y^2} + 2.y.1 + {1^2}} \right) = 0\\
 \Leftrightarrow {\left( {x - 2} \right)^2} + 3.{\left( {y + 1} \right)^2} = 0\,\,\,\,\,\left( 1 \right)\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y + 1} \right)^2} \ge 0,\,\,\,\forall y\\
 \Rightarrow {\left( {x - 2} \right)^2} + 3{\left( {y + 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} = 0\\
{\left( {y + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 2 = 0\\
y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y =  - 1
\end{array} \right.\\
4)\,\,\,3{x^2} + {y^2} + 10x - 2xy + 26 = 0\\
 \Leftrightarrow \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{x^2} + 10x + 25} \right) + {x^2} = 0\\
 \Leftrightarrow {\left( {x - y} \right)^2} + \left( {{x^2} + 2.x.5 + {5^2}} \right) + {x^2} = 0\\
 \Leftrightarrow {\left( {x - y} \right)^2} + {\left( {x + 5} \right)^2} + {x^2} = 0\,\,\,\,\left( 2 \right)\\
{\left( {x - y} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x + 5} \right)^2} \ge 0,\,\,\,\forall x\\
{x^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow {\left( {x - y} \right)^2} + {\left( {x + 5} \right)^2} + {x^2} \ge 0\,,\,\,\,\forall x,y\\
\left( 2 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {x + 5} \right)^2} = 0\\
{x^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
x + 5 = 0\\
x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = y\\
x =  - 5\\
x = 0
\end{array} \right.\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
5)\,\,\,3{x^2} + 6{y^2} - 12x - 20y + 40 = 0\\
 \Leftrightarrow \left( {3{x^2} - 12x + 12} \right) + \left( {6{y^2} - 20y + \dfrac{{50}}{3}} \right) + \dfrac{{34}}{3} = 0\\
 \Leftrightarrow 3.\left( {{x^2} - 4x + 4} \right) + 6.\left( {{y^2} - \dfrac{{10}}{3}y + \dfrac{{25}}{9}} \right) + \dfrac{{34}}{3} = 0\\
 \Leftrightarrow 3.\left( {{x^2} - 2.x.2 + {2^2}} \right) + 6.\left[ {{y^2} - 2.y.\dfrac{5}{3} + {{\left( {\dfrac{5}{3}} \right)}^2}} \right] + \dfrac{{34}}{3} = 0\\
 \Leftrightarrow 3.{\left( {x - 2} \right)^2} + 6.{\left( {y - \dfrac{5}{3}} \right)^2} + \dfrac{{34}}{3} = 0\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - \dfrac{5}{3}} \right)^2} \ge 0,\,\,\,\forall y\\
 \Rightarrow 3.{\left( {x - 2} \right)^2} + 6.{\left( {y - \dfrac{5}{3}} \right)^2} + \dfrac{{34}}{3} \ge \dfrac{{34}}{3} > 0,\,\,\,\forall x,y\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm
\end{array}\)