Giải pt: a.($2x^{2}$-3)$^{2}$ – 4($x-1)^{2}$ =0 b, 2x($3x-1)^{2}$ – $9x^{2}$ +1=0 c. $3x^{2}$ -14|x|-5=0

Đáp án:

c) \(\left[ \begin{array}{l}
x = 5\\
x =  – 5
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a){\left( {2{x^2} – 3} \right)^2} = 4{\left( {x – 1} \right)^2}\\
 \to \left| {2{x^2} – 3} \right| = 2\left| {x – 1} \right|\\
 \to \left[ \begin{array}{l}
2x – 2 = 2{x^2} – 3\left( {DK:x > 1} \right)\\
2x – 2 =  – 2{x^2} + 3\left( {DK:x \le 1} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2{x^2} – 2x – 1 = 0\\
2{x^2} + 3x – 5 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt 3 }}{2}\\
x = \dfrac{{1 – \sqrt 3 }}{2}\left( l \right)\\
x = 1\\
x =  – \dfrac{5}{2}
\end{array} \right.\\
b)2x{\left( {3x – 1} \right)^2} – \left( {9{x^2} – 1} \right) = 0\\
 \to 2x{\left( {3x – 1} \right)^2} – \left( {3x – 1} \right)\left( {3x + 1} \right) = 0\\
 \to \left( {3x – 1} \right)\left[ {2x\left( {3x – 1} \right) – 3x – 1} \right] = 0\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
6{x^2} – 2x – 3x – 1 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = 1\\
x =  – \dfrac{1}{6}
\end{array} \right.\\
c)3{x^2} – 14\left| x \right| – 5 = 0\\
 \to \left[ \begin{array}{l}
\left| x \right| = 5\\
\left| x \right| =  – \dfrac{1}{3}\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  – 5
\end{array} \right.
\end{array}\)