Trang chủ » Hỏi đáp » Môn Toán Bài 1: tìm x, biết a)x^3-4x=2-x b) x^2(x – 6)=x^2 – 36 c)x^3 +x- x^2 – 1= 0 mong mn giúp mần;-; 15/01/2025 Bài 1: tìm x, biết a)x^3-4x=2-x b) x^2(x – 6)=x^2 – 36 c)x^3 +x- x^2 – 1= 0 mong mn giúp mần;-;
Giải đáp: $\begin{array}{l}a)x = 2;x = – 1\\b)x = – 2;x = 3;x = 6\\c)x = 1\end{array}$ Lời giải và giải thích chi tiết: $\begin{array}{l}a){x^3} – 4x = 2 – x\\ \Leftrightarrow x\left( {{x^2} – 4} \right) + x – 2 = 0\\ \Leftrightarrow x\left( {x + 2} \right)\left( {x – 2} \right) + x – 2 = 0\\ \Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 1} \right) = 0\\ \Leftrightarrow \left( {x – 2} \right){\left( {x + 1} \right)^2} = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 2 = 0\\x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\\x = – 1\end{array} \right.\\Vậy\,x = 2;x = – 1\\b){x^2}\left( {x – 6} \right) = {x^2} – 36\\ \Leftrightarrow {x^2}\left( {x – 6} \right) = \left( {x – 6} \right)\left( {x + 6} \right)\\ \Leftrightarrow {x^2}\left( {x – 6} \right) – \left( {x – 6} \right)\left( {x + 6} \right) = 0\\ \Leftrightarrow \left( {x – 6} \right)\left( {{x^2} – x – 6} \right) = 0\\ \Leftrightarrow \left( {x – 6} \right)\left( {x – 3} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 6\\x = 3\\x = – 2\end{array} \right.\\Vậy\,x = – 2;x = 3;x = 6\\c){x^3} + x – {x^2} – 1 = 0\\ \Leftrightarrow x\left( {{x^2} + 1} \right) – \left( {{x^2} + 1} \right) = 0\\ \Leftrightarrow \left( {{x^2} + 1} \right)\left( {x – 1} \right) = 0\\ \Leftrightarrow x – 1 = 0\\ \Leftrightarrow x = 1\\Vậy\,x = 1\end{array}$ Trả lời
a)x = 2;x = – 1\\
b)x = – 2;x = 3;x = 6\\
c)x = 1
\end{array}$
a){x^3} – 4x = 2 – x\\
\Leftrightarrow x\left( {{x^2} – 4} \right) + x – 2 = 0\\
\Leftrightarrow x\left( {x + 2} \right)\left( {x – 2} \right) + x – 2 = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 1} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right){\left( {x + 1} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.\\
Vậy\,x = 2;x = – 1\\
b){x^2}\left( {x – 6} \right) = {x^2} – 36\\
\Leftrightarrow {x^2}\left( {x – 6} \right) = \left( {x – 6} \right)\left( {x + 6} \right)\\
\Leftrightarrow {x^2}\left( {x – 6} \right) – \left( {x – 6} \right)\left( {x + 6} \right) = 0\\
\Leftrightarrow \left( {x – 6} \right)\left( {{x^2} – x – 6} \right) = 0\\
\Leftrightarrow \left( {x – 6} \right)\left( {x – 3} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = 3\\
x = – 2
\end{array} \right.\\
Vậy\,x = – 2;x = 3;x = 6\\
c){x^3} + x – {x^2} – 1 = 0\\
\Leftrightarrow x\left( {{x^2} + 1} \right) – \left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 1} \right)\left( {x – 1} \right) = 0\\
\Leftrightarrow x – 1 = 0\\
\Leftrightarrow x = 1\\
Vậy\,x = 1
\end{array}$