Đáp án:
B5:
Min=4
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a){x^2} - 1 - {x^2} - x = 4\\
\to - x = 5\\
\to x = - 5\\
b)3x\left( {x - 2015} \right) - \left( {x - 2015} \right) = 0\\
\to \left( {x - 2015} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2015\\
x = \dfrac{1}{3}
\end{array} \right.\\
c)x\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
d){x^2} + 6x + 9 = 0\\
\to {\left( {x + 3} \right)^2} = 0\\
\to x + 3 = 0\\
\to x = - 3\\
B5:\\
A = {x^2} - 6x + 9 + 4\\
= {\left( {x - 3} \right)^2} + 4\\
Do:{\left( {x - 3} \right)^2} \ge 0\\
\to {\left( {x - 3} \right)^2} + 4 \ge 4\\
\to Min = 4\\
\Leftrightarrow x = 3
\end{array}\)
