Tìm x: (x+3)(x^2-3x+9)-x (x+2)(x-2)=15 4x^2-8x+3=0 03/10/2021 Bởi Ayla Tìm x: (x+3)(x^2-3x+9)-x (x+2)(x-2)=15 4x^2-8x+3=0
\[\begin{array}{l} \left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) – x\left( {x + 2} \right)\left( {x – 2} \right) = 15\\ \Leftrightarrow {x^3} + 27 – x\left( {{x^2} – 4} \right) = 15\\ \Leftrightarrow {x^3} + 27 – {x^3} + 4x = 15\\ \Leftrightarrow 4x = – 12\\ \Leftrightarrow x = – 3.\\ 4{x^2} – 8x + 3 = 0\\ \Leftrightarrow 4{x^2} – 2x – 6x + 3 = 0\\ \Leftrightarrow 2x\left( {2x – 1} \right) – 3\left( {2x – 1} \right) = 0\\ \Leftrightarrow \left( {2x – 1} \right)\left( {2x – 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{1}{2}\\ x = \frac{3}{2} \end{array} \right.. \end{array}\] Bình luận
Đáp án: câu1:x=-3
Câu2:x=0.5
Giải thích các bước giải:
\[\begin{array}{l}
\left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) – x\left( {x + 2} \right)\left( {x – 2} \right) = 15\\
\Leftrightarrow {x^3} + 27 – x\left( {{x^2} – 4} \right) = 15\\
\Leftrightarrow {x^3} + 27 – {x^3} + 4x = 15\\
\Leftrightarrow 4x = – 12\\
\Leftrightarrow x = – 3.\\
4{x^2} – 8x + 3 = 0\\
\Leftrightarrow 4{x^2} – 2x – 6x + 3 = 0\\
\Leftrightarrow 2x\left( {2x – 1} \right) – 3\left( {2x – 1} \right) = 0\\
\Leftrightarrow \left( {2x – 1} \right)\left( {2x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{1}{2}\\
x = \frac{3}{2}
\end{array} \right..
\end{array}\]