Toán |x^2-1|=2x+1 giúp em với các ad toán học 17/10/2021 By Brielle |x^2-1|=2x+1 giúp em với các ad toán học
Đáp án: \(\left[ \begin{array}{l}x = 1 + \sqrt 3 \\x = 0\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}Xét:{x^2} – 1 > 0\\ \to \left( {x – 1} \right)\left( {x + 1} \right) > 0\\ \to \left[ \begin{array}{l}\left\{ \begin{array}{l}x – 1 > 0\\x + 1 > 0\end{array} \right.\\\left\{ \begin{array}{l}x – 1 < 0\\x + 1 < 0\end{array} \right.\end{array} \right. \to \left[ \begin{array}{l}x > 1\\x < – 1\end{array} \right.\\{x^2} – 1 \le 0 \to – 1 \le x \le 1\\Có:\left| {{x^2} – 1} \right| = 2x + 1\\ \to \left[ \begin{array}{l}{x^2} – 1 = 2x + 1\left( {DK:\left[ \begin{array}{l}x > 1\\x < – 1\end{array} \right.} \right)\\{x^2} – 1 = – 2x – 1\left( {DK: – 1 \le x \le 1} \right)\end{array} \right.\\ \to \left[ \begin{array}{l}{x^2} – 2x – 2 = 0\left( 1 \right)\\{x^2} + 2x = 0\left( 2 \right)\end{array} \right.\\Pt\left( 1 \right): Δ’= 1 + 2 = 3\\ \to \left[ \begin{array}{l}x = 1 + \sqrt 3 \left( {TM} \right)\\x = 1 – \sqrt 3 \left( l \right)\end{array} \right.\\Pt\left( 2 \right):x\left( {x + 2} \right) = 0\\ \to \left[ \begin{array}{l}x = 0\left( {TM} \right)\\x = – 2\left( l \right)\end{array} \right.\\KL:\left[ \begin{array}{l}x = 1 + \sqrt 3 \\x = 0\end{array} \right.\end{array}\) Trả lời
Đáp án:
\(\left[ \begin{array}{l}
x = 1 + \sqrt 3 \\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:{x^2} – 1 > 0\\
\to \left( {x – 1} \right)\left( {x + 1} \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 1 > 0\\
x + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 1 < 0\\
x + 1 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x > 1\\
x < – 1
\end{array} \right.\\
{x^2} – 1 \le 0 \to – 1 \le x \le 1\\
Có:\left| {{x^2} – 1} \right| = 2x + 1\\
\to \left[ \begin{array}{l}
{x^2} – 1 = 2x + 1\left( {DK:\left[ \begin{array}{l}
x > 1\\
x < – 1
\end{array} \right.} \right)\\
{x^2} – 1 = – 2x – 1\left( {DK: – 1 \le x \le 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} – 2x – 2 = 0\left( 1 \right)\\
{x^2} + 2x = 0\left( 2 \right)
\end{array} \right.\\
Pt\left( 1 \right): Δ’= 1 + 2 = 3\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt 3 \left( {TM} \right)\\
x = 1 – \sqrt 3 \left( l \right)
\end{array} \right.\\
Pt\left( 2 \right):x\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = – 2\left( l \right)
\end{array} \right.\\
KL:\left[ \begin{array}{l}
x = 1 + \sqrt 3 \\
x = 0
\end{array} \right.
\end{array}\)