Giải thích các bước giải:
\(\begin{array}{l}
1,\\
a,\\
{\left( {x + 4} \right)^2} + \left( {1 - x} \right)\left( {1 + x} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 8x + 16} \right) + \left( {1 - {x^2}} \right) = 0\\
\Leftrightarrow 8x + 17 = 0\\
\Leftrightarrow x = - \dfrac{{17}}{8}\\
b,\\
{x^2} - {\left( {x + 5} \right)^2} = 0\\
\Leftrightarrow \left[ {x - \left( {x + 5} \right)} \right].\left[ {x + \left( {x + 5} \right)} \right] = 0\\
\Leftrightarrow \left( { - 5} \right).\left( {2x + 5} \right) = 0\\
\Leftrightarrow 2x + 5 = 0\\
\Leftrightarrow x = - \dfrac{5}{2}\\
c,\\
{\left( {4x + 1} \right)^2} - {\left( {4x} \right)^2} = 0\\
\Leftrightarrow \left[ {\left( {4x + 1} \right) - 4x} \right].\left[ {\left( {4x + 1} \right) + 4x} \right] = 0\\
\Leftrightarrow 1.\left( {8x + 1} \right) = 0\\
\Leftrightarrow 8x + 1 = 0\\
\Leftrightarrow x = - \dfrac{1}{8}\\
d,\\
4{x^2} - 7x + 3 = 0\\
\Leftrightarrow \left( {4{x^2} - 4x} \right) - \left( {3x - 3} \right) = 0\\
\Leftrightarrow 4x.\left( {x - 1} \right) - 3.\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {4x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
4x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{4}
\end{array} \right.\\
2,\\
{x^2} + {y^2} = \left( {{x^2} + 2xy + {y^2}} \right) - 2xy = {\left( {x + y} \right)^2} - 2xy\\
= {10^2} - 2.30 = 100 - 60 = 40\\
{\left( {x - y} \right)^2} = {x^2} - 2xy + {y^2} = \left( {{x^2} + {y^2}} \right) - 2xy = 40 - 2.30 = - 20\\
{\left( {x - y} \right)^2} \ge 0 \Rightarrow vn
\end{array}\)
