Đáp án: 5) \(x = \dfrac{{15}}{2}\) Giải thích các bước giải: \(\begin{array}{l}1)5\left( {4{x^2} – 12x + 9} \right) – 5\left( {{x^2} + 2x + 1} \right) – 15\left( {{x^2} – 16} \right) = – 10\\ \to 20{x^2} – 60x + 45 – 5{x^2} – 10x – 5 – 15{x^2} + 240 + 10 = 0\\ \to – 70x + 290 = 0\\ \to x = \dfrac{{29}}{7}\\2)x\left( {{x^2} – 25} \right) – {x^3} + 2{x^2} – 4x – 2{x^2} + 4x – 8 = 3\\ \to {x^3} – 25x – {x^3} + 2{x^2} – 4x – 2{x^2} + 4x – 8 = 3\\ \to – 25x = 11\\ \to x = – \dfrac{{11}}{{25}}\\3){x^3} – 3{x^2} + 3x – 1 + 8 + 4x + 2{x^2} – 4x – 2{x^2} – {x^3} + 3{x^2} + 6 = 17\\ \to 3x = 4\\ \to x = \dfrac{4}{3}\\4){x^3} – 2{x^2} + 4x + 2{x^2} – 4x + 8 – {x^3} – 2{x^2} = 5\\ \to – 2{x^2} = – 3\\ \to 2{x^2} = 3\\ \to \left[ \begin{array}{l}x = \sqrt {\dfrac{3}{2}} \\x = – \sqrt {\dfrac{3}{2}} \end{array} \right.\\5){x^2} + 4x + 4 + {x^2} – 6x + 9 – 2\left( {{x^2} – 1} \right) = 0\\ \to 2{x^2} – 2x + 13 – 2{x^2} + 2 = 0\\ \to – 2x = – 15\\ \to x = \dfrac{{15}}{2}\end{array}\) Reply
Đáp án:
5) \(x = \dfrac{{15}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)5\left( {4{x^2} – 12x + 9} \right) – 5\left( {{x^2} + 2x + 1} \right) – 15\left( {{x^2} – 16} \right) = – 10\\
\to 20{x^2} – 60x + 45 – 5{x^2} – 10x – 5 – 15{x^2} + 240 + 10 = 0\\
\to – 70x + 290 = 0\\
\to x = \dfrac{{29}}{7}\\
2)x\left( {{x^2} – 25} \right) – {x^3} + 2{x^2} – 4x – 2{x^2} + 4x – 8 = 3\\
\to {x^3} – 25x – {x^3} + 2{x^2} – 4x – 2{x^2} + 4x – 8 = 3\\
\to – 25x = 11\\
\to x = – \dfrac{{11}}{{25}}\\
3){x^3} – 3{x^2} + 3x – 1 + 8 + 4x + 2{x^2} – 4x – 2{x^2} – {x^3} + 3{x^2} + 6 = 17\\
\to 3x = 4\\
\to x = \dfrac{4}{3}\\
4){x^3} – 2{x^2} + 4x + 2{x^2} – 4x + 8 – {x^3} – 2{x^2} = 5\\
\to – 2{x^2} = – 3\\
\to 2{x^2} = 3\\
\to \left[ \begin{array}{l}
x = \sqrt {\dfrac{3}{2}} \\
x = – \sqrt {\dfrac{3}{2}}
\end{array} \right.\\
5){x^2} + 4x + 4 + {x^2} – 6x + 9 – 2\left( {{x^2} – 1} \right) = 0\\
\to 2{x^2} – 2x + 13 – 2{x^2} + 2 = 0\\
\to – 2x = – 15\\
\to x = \dfrac{{15}}{2}
\end{array}\)