Toán Lớp 8: (x^2-3)^2-(2x-1)^2=0
lm hộ mik với ạ
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Toán Lớp 8: (x^2-3)^2-(2x-1)^2=0
lm hộ mik với ạ
Comments ( 1 )
x = 1 + \sqrt 3 \\
x = – 1 + \sqrt 5
\end{array} \right.\)
{({x^2} – 3)^2} – {(2x – 1)^2} = 0\\
\to {({x^2} – 3)^2} = {(2x – 1)^2}\\
\to \left| {{x^2} – 3} \right| = \left| {2x – 1} \right|\\
\to \left[ \begin{array}{l}
{x^2} – 3 = 2x – 1\left( {DK:\left[ \begin{array}{l}
x \ge \sqrt 3 \\
x \le – \sqrt 3
\end{array} \right.} \right)\\
{x^2} – 3 = – 2x + 1\left( {DK:\sqrt 3 > x > – \sqrt 3 } \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} – 2x – 2 = 0\\
{x^2} + 2x – 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt 3 \\
x = 1 – \sqrt 3 \left( l \right)\\
x = – 1 + \sqrt 5 \\
x = – 1 – \sqrt 5 \left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt 3 \\
x = – 1 + \sqrt 5
\end{array} \right.
\end{array}\)