a)1/2x ×(x^2-4)=0 b)x × (x+1)-x × (x-3) c)(x-2)×(x+4)-(x+1)^2 d)4x^2+16x+16 29/11/2021 Bởi Melanie a)1/2x ×(x^2-4)=0 b)x × (x+1)-x × (x-3) c)(x-2)×(x+4)-(x+1)^2 d)4x^2+16x+16
Đáp án: d) \(4{\left( {x + 2} \right)^2}\) Giải thích các bước giải: \(\begin{array}{l}a)\dfrac{1}{2}x\left( {{x^2} – 4} \right) = 0\\ \to \left[ \begin{array}{l}x = 0\\\left( {x – 2} \right)\left( {x + 2} \right) = 0\end{array} \right.\\ \to \left[ \begin{array}{l}x = 0\\x = 2\\x = – 2\end{array} \right.\\b)x\left( {x + 1} \right) – x\left( {x – 3} \right) = 0\\ \to x\left( {x + 1 – x + 3} \right) = 0\\ \to 4x = 0\\ \to x = 0\\c)\left( {x – 2} \right)\left( {x + 4} \right) – {\left( {x + 1} \right)^2}\\ = {x^2} + 2x – 8 – {x^2} – 2x – 1\\ = – 9\\d)4{x^2} + 16x + 16\\ = 4\left( {{x^2} + 4x + 4} \right)\\ = 4{\left( {x + 2} \right)^2}\end{array}\) Bình luận
Đáp án:
d) \(4{\left( {x + 2} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{1}{2}x\left( {{x^2} – 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
\left( {x – 2} \right)\left( {x + 2} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = – 2
\end{array} \right.\\
b)x\left( {x + 1} \right) – x\left( {x – 3} \right) = 0\\
\to x\left( {x + 1 – x + 3} \right) = 0\\
\to 4x = 0\\
\to x = 0\\
c)\left( {x – 2} \right)\left( {x + 4} \right) – {\left( {x + 1} \right)^2}\\
= {x^2} + 2x – 8 – {x^2} – 2x – 1\\
= – 9\\
d)4{x^2} + 16x + 16\\
= 4\left( {{x^2} + 4x + 4} \right)\\
= 4{\left( {x + 2} \right)^2}
\end{array}\)