Đáp án:

\(\begin{array}{l}
Bài\,\,\,5:\,\\
1,\\
{\left( {x + 5} \right)^2} + {\left( {y + 1} \right)^2}\\
2,\\
{\left( {z - 3} \right)^2} + {\left( {t + 2} \right)^2}\\
3,\\
{\left( {x - y} \right)^2} + {\left( {y + 1} \right)^2}\\
4,\\
{\left( {2x - z} \right)^2} + {\left( {z - 1} \right)^2}\\
5,\\
{\left( {2x - 3} \right)^2} - {\left( {y - 1} \right)^2}\\
6,\\
{\left( {2x - z} \right)^2} + {\left( {z - 1} \right)^2}\\
7,\\
{\left( {x + y} \right)^2} - {4^2}\\
8,\\
{x^2} - {\left( {y - 6} \right)^2}\\
9,\\
{\left( {y - 3} \right)^2} - {\left( {2z} \right)^2}\\
10,\\
{\left( {2y + 3z} \right)^2} - {x^2}\\
Bài\,\,\,7:\\
1,\\
\left[ \begin{array}{l}
x = \frac{3}{5}\\
x =  - \frac{3}{5}
\end{array} \right.\\
2,\\
\left[ \begin{array}{l}
x = 5\\
x = 1
\end{array} \right.\\
3,\\
\left[ \begin{array}{l}
x = 6\\
x =  - 4
\end{array} \right.\\
4,\\
x =  - \frac{1}{8}\\
5,\\
x =  - \frac{{255}}{2}\\
6,\\
x = \frac{{17}}{6}\\
7,\\
x = 0\\
8,\\
x = 1\\
9,\\
x = \frac{3}{2}\\
10,\\
x = 14
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
Bài\,\,\,5:\,\\
1,\\
{x^2} + 10x + 26 + {y^2} + 2y\\
 = \left( {{x^2} + 10x + 25} \right) + \left( {{y^2} + 2y + 1} \right)\\
 = \left( {{x^2} + 2.x.5 + {5^2}} \right) + \left( {{y^2} + 2.y.1 + {1^2}} \right)\\
 = {\left( {x + 5} \right)^2} + {\left( {y + 1} \right)^2}\\
2,\\
{z^2} - 6z + 13 + {t^2} + 4t\\
 = \left( {{z^2} - 6z + 9} \right) + \left( {{t^2} + 4t + 4} \right)\\
 = \left( {{z^2} - 2.z.3 + {3^2}} \right) + \left( {{t^2} + 2.t.2 + {2^2}} \right)\\
 = {\left( {z - 3} \right)^2} + {\left( {t + 2} \right)^2}\\
3,\\
{x^2} - 2xy + 2{y^2} + 2y + 1\\
 = \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{y^2} + 2y + 1} \right)\\
 = \left( {{x^2} - 2.x.y + {y^2}} \right) + \left( {{y^2} + 2.y.1 + {1^2}} \right)\\
 = {\left( {x - y} \right)^2} + {\left( {y + 1} \right)^2}\\
4,\\
4{x^2} + 2{z^2} - 4xz - 2z + 1\\
 = \left( {4{x^2} - 4xz + {z^2}} \right) + \left( {{z^2} - 2z + 1} \right)\\
 = \left[ {{{\left( {2x} \right)}^2} - 2.2x.z + {z^2}} \right] + \left( {{z^2} - 2.z.1 + {1^2}} \right)\\
 = {\left( {2x - z} \right)^2} + {\left( {z - 1} \right)^2}\\
5,\\
4{x^2} - 12x - {y^2} + 2y + 8\\
 = \left( {4{x^2} - 12x + 9} \right) + \left( { - {y^2} + 2y - 1} \right)\\
 = \left[ {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}} \right] - \left( {{y^2} - 2y + 1} \right)\\
 = {\left( {2x - 3} \right)^2} - {\left( {y - 1} \right)^2}\\
6,\\
4{x^2} + 2{z^2} - 4zx - 2z + 1\\
 = \left( {4{x^2} - 4zx + {z^2}} \right) + \left( {{z^2} - 2z + 1} \right)\\
 = \left[ {{{\left( {2x} \right)}^2} - 2.2x.z + {z^2}} \right] + \left( {{z^2} - 2.z.1 + {1^2}} \right)\\
 = {\left( {2x - z} \right)^2} + {\left( {z - 1} \right)^2}\\
7,\\
\left( {x + y + 4} \right).\left( {x + y - 4} \right)\\
 = \left[ {\left( {x + y} \right) + 4} \right].\left[ {\left( {x + y} \right) - 4} \right]\\
 = {\left( {x + y} \right)^2} - {4^2}\\
8,\\
\left( {x - y + 6} \right).\left( {x + y - 6} \right)\\
 = \left[ {x - \left( {y - 6} \right)} \right].\left[ {x + \left( {y - 6} \right)} \right]\\
 = {x^2} - {\left( {y - 6} \right)^2}\\
9,\\
\left( {y + 2z - 3} \right).\left( {y - 2z - 3} \right)\\
 = \left[ {\left( {y - 3} \right) + 2z} \right].\left[ {\left( {y - 3} \right) - 2z} \right]\\
 = {\left( {y - 3} \right)^2} - {\left( {2z} \right)^2}\\
10,\\
\left( {x + 2y + 3z} \right).\left( {2y + 3z - x} \right)\\
 = \left[ {\left( {2y + 3z} \right) + x} \right].\left[ {\left( {2y + 3z} \right) - x} \right]\\
 = {\left( {2y + 3z} \right)^2} - {x^2}\\
Bài\,\,\,7:\\
1,\\
25{x^2} - 9 = 0\\
 \Leftrightarrow {\left( {5x} \right)^2} - {3^2} = 0\\
 \Leftrightarrow \left( {5x - 3} \right)\left( {5x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
5x - 3 = 0\\
5x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{3}{5}\\
x =  - \frac{3}{5}
\end{array} \right.\\
2,\\
{\left( {x - 3} \right)^2} - 4 = 0\\
 \Leftrightarrow {\left( {x - 3} \right)^2} - {2^2} = 0\\
 \Leftrightarrow \left[ {\left( {x - 3} \right) - 2} \right].\left[ {\left( {x - 3} \right) + 2} \right] = 0\\
 \Leftrightarrow \left( {x - 5} \right).\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 1
\end{array} \right.\\
3,\\
{x^2} - 2x = 24\\
 \Leftrightarrow {x^2} - 2x + 1 = 25\\
 \Leftrightarrow \left( {{x^2} - 2.x.1 + {1^2}} \right) - 25 = 0\\
 \Leftrightarrow {\left( {x - 1} \right)^2} - {5^2} = 0\\
 \Leftrightarrow \left[ {\left( {x - 1} \right) - 5} \right].\left[ {\left( {x - 1} \right) + 5} \right] = 0\\
 \Leftrightarrow \left( {x - 6} \right).\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 6 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x =  - 4
\end{array} \right.\\
4,\\
{\left( {x + 4} \right)^2} - \left( {x + 1} \right).\left( {x - 1} \right) = 16\\
 \Leftrightarrow \left( {{x^2} + 2.x.4 + {4^2}} \right) - \left( {{x^2} - {1^2}} \right) = 16\\
 \Leftrightarrow \left( {{x^2} + 8x + 16} \right) - \left( {{x^2} - 1} \right) = 16\\
 \Leftrightarrow {x^2} + 8x + 16 - {x^2} + 1 = 16\\
 \Leftrightarrow 8x + 1 = 0\\
 \Leftrightarrow x =  - \frac{1}{8}\\
5,\\
{\left( {2x - 1} \right)^2} + {\left( {x + 3} \right)^2} - 5.\left( {x + 7} \right).\left( {x - 7} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - 2.2x.1 + {1^2}} \right] + \left( {{x^2} + 2.x.3 + {3^2}} \right) - 5.\left( {{x^2} - {7^2}} \right) = 0\\
 \Leftrightarrow \left( {4{x^2} - 4x + 1} \right) + \left( {{x^2} + 6x + 9} \right) - 5.\left( {{x^2} - 49} \right) = 0\\
 \Leftrightarrow 4{x^2} - 4x + 1 + {x^2} + 6x + 9 - 5{x^2} + 245 = 0\\
 \Leftrightarrow 2x + 255 = 0\\
 \Leftrightarrow x =  - \frac{{255}}{2}\\
6,\\
3{\left( {x - 1} \right)^2} - 3x\left( {x - 5} \right) = 1\\
 \Leftrightarrow 3.\left( {{x^2} - 2.x.1 + {1^2}} \right) - \left( {3{x^2} - 15} \right) = 1\\
 \Leftrightarrow 3\left( {{x^2} - 2x + 1} \right) - 3{x^2} + 15 = 1\\
 \Leftrightarrow 3{x^2} - 6x + 3 - 3{x^2} + 15 - 1 = 0\\
 \Leftrightarrow  - 6x + 17 = 0\\
 \Leftrightarrow x = \frac{{17}}{6}\\
7,\\
{\left( {6x - 2} \right)^2} + {\left( {5x - 2} \right)^2} - 4.\left( {3x - 1} \right).\left( {5x - 2} \right) = 0\\
 \Leftrightarrow {\left( {6x - 2} \right)^2} + {\left( {5x - 2} \right)^2} - 2.2\left( {3x - 1} \right).\left( {5x - 2} \right) = 0\\
 \Leftrightarrow {\left( {6x - 2} \right)^2} + {\left( {5x - 2} \right)^2} - 2.\left( {6x - 2} \right).\left( {5x - 2} \right) = 0\\
 \Leftrightarrow {\left( {6x - 2} \right)^2} - 2.\left( {6x - 2} \right).\left( {5x - 2} \right) + {\left( {5x - 2} \right)^2} = 0\\
 \Leftrightarrow {\left[ {\left( {6x - 2} \right) - \left( {5x - 2} \right)} \right]^2} = 0\\
 \Leftrightarrow {\left( {6x - 2 - 5x + 2} \right)^2} = 0\\
 \Leftrightarrow {x^2} = 0\\
 \Leftrightarrow x = 0\\
8,\\
{\left( {x - 2} \right)^3} - {x^2}\left( {x - 6} \right) = 4\\
 \Leftrightarrow \left( {{x^3} - 3.{x^2}.2 + 3.x{{.2}^2} - {2^3}} \right) - \left( {{x^3} - 6{x^2}} \right) = 4\\
 \Leftrightarrow \left( {{x^3} - 6{x^2} + 12x - 8} \right) - \left( {{x^3} - 6{x^2}} \right) = 4\\
 \Leftrightarrow {x^3} - 6{x^2} + 12x - 8 - {x^3} + 6{x^2} - 4 = 0\\
 \Leftrightarrow 12x - 12 = 0\\
 \Leftrightarrow x = 1\\
9,\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) - x\left( {x + 2} \right)\left( {x - 2} \right) = 5\\
 \Leftrightarrow \left( {x - 1} \right).\left( {{x^2} + x.1 + {1^2}} \right) - x.\left( {{x^2} - {2^2}} \right) = 5\\
 \Leftrightarrow \left( {{x^3} - {1^3}} \right) - x.\left( {{x^2} - 4} \right) = 5\\
 \Leftrightarrow \left( {{x^3} - 1} \right) - \left( {{x^3} - 4x} \right) - 5 = 0\\
 \Leftrightarrow {x^3} - 1 - {x^3} + 4x - 5 = 0\\
 \Leftrightarrow 4x - 6 = 0\\
 \Leftrightarrow x = \frac{3}{2}\\
10,\\
{\left( {x - 1} \right)^3} - \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) + 3.\left( {{x^2} - 4} \right) = 2\\
 \Leftrightarrow \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - \left( {x + 3} \right).\left( {{x^2} - x.3 + {3^2}} \right) + 3{x^2} - 12 = 2\\
 \Leftrightarrow \left( {{x^3} - 3{x^2} + 3x - 1} \right) - \left( {{x^3} + {3^3}} \right) + 3{x^2} - 12 - 2 = 0\\
 \Leftrightarrow {x^3} - 3{x^2} + 3x - 1 - {x^3} - 27 + 3{x^2} - 14 = 0\\
 \Leftrightarrow 3x - 42 = 0\\
 \Leftrightarrow x = 14
\end{array}\)