Giải phương trình: 2x^5- 3x^4 – 5x^3 + 5x^2 + 3x – 2=0 20/08/2021 Bởi Julia Giải phương trình: 2x^5- 3x^4 – 5x^3 + 5x^2 + 3x – 2=0
Đáp án: \(\left[ \begin{array}{l}x = 1\\x = 2\\x = – 1\\x = \frac{1}{2}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}2{x^5} – 3{x^4} – 5{x^3} + 5{x^2} + 3x – 2 = 0\\ \leftrightarrow 2{x^5} – 2{x^4} – {x^4} + {x^3} – 6{x^3} + 6{x^2} – {x^2} + x + 2x – 2 = 0\\ \leftrightarrow 2{x^4}(x – 1) – {x^3}(x – 1) – 6{x^2}(x – 1) – x(x – 1) + 2(x – 1) = 0\\ \leftrightarrow (x – 1)(2{x^4} – {x^3} – 6{x^2} – x + 2) = 0\\ \leftrightarrow (x – 1)(2{x^4} – 4{x^3} + 3{x^3} – 6{x^2} – x + 2) = 0\\ \leftrightarrow (x – 1)\left[ {2{x^3}(x – 2) + 3{x^2}(x – 2) – (x – 2)} \right] = 0\\ \leftrightarrow (x – 1)(x – 2)(2{x^3} + 3{x^2} – 1) = 0\\ \leftrightarrow \left[ \begin{array}{l}x = 1\\x = 2\\2{x^3} + 3{x^2} – 1 = 0\end{array} \right. \leftrightarrow \left[ \begin{array}{l}x = 1\\x = 2\\x = – 1\\x = \frac{1}{2}\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x = 1\\
x = 2\\
x = – 1\\
x = \frac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
2{x^5} – 3{x^4} – 5{x^3} + 5{x^2} + 3x – 2 = 0\\
\leftrightarrow 2{x^5} – 2{x^4} – {x^4} + {x^3} – 6{x^3} + 6{x^2} – {x^2} + x + 2x – 2 = 0\\
\leftrightarrow 2{x^4}(x – 1) – {x^3}(x – 1) – 6{x^2}(x – 1) – x(x – 1) + 2(x – 1) = 0\\
\leftrightarrow (x – 1)(2{x^4} – {x^3} – 6{x^2} – x + 2) = 0\\
\leftrightarrow (x – 1)(2{x^4} – 4{x^3} + 3{x^3} – 6{x^2} – x + 2) = 0\\
\leftrightarrow (x – 1)\left[ {2{x^3}(x – 2) + 3{x^2}(x – 2) – (x – 2)} \right] = 0\\
\leftrightarrow (x – 1)(x – 2)(2{x^3} + 3{x^2} – 1) = 0\\
\leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
2{x^3} + 3{x^2} – 1 = 0
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = – 1\\
x = \frac{1}{2}
\end{array} \right.
\end{array}\)