a)
pt <=> \(x^2=324\)
<=> \(\orbr{\begin{cases}x=18\\x=-18\end{cases}}\)
Vậy tập hợp nghiệm của pt là: S={18; -18}
b) pt <=> \(16x^2=5\)
<=> \(x^2=\frac{5}{16}\)
<=> \(\orbr{\begin{cases}x=\frac{\sqrt{5}}{4}\\x=-\frac{\sqrt{5}}{4}\end{cases}}\)
a. \(-x^2+324=0\)
\(\Leftrightarrow-x^2=-324\)
\(\Leftrightarrow x^2=324=18^2\)
\(\Leftrightarrow x=18;x=-18\)
b. \(16x^2-5=0\)
\(\Leftrightarrow16x^2=5\)
\(\Leftrightarrow x^2=\frac{5}{16}=\frac{\sqrt{5}}{4}^2\)
\(\Leftrightarrow x=\frac{\sqrt{5}}{4}\)
c) pt <=> \(4\sqrt{x-3}=2\)
<=> \(2\sqrt{x-3}=1\)
<=> \(4\left(x-3\right)=1\)
<=> \(4x-12=1\)
<=> \(x=\frac{13}{4}\)
d) pt <=> \(4x^2-4x+1=9\)
<=> \(\left(2x-1\right)^2=9\)
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
a, \(-x^2+324=0\Leftrightarrow-x^2=-324\Leftrightarrow x^2=324\Leftrightarrow x=\pm18\)
b, \(16x^2-5=0\Leftrightarrow16x^2=5\Leftrightarrow x^2=\frac{5}{16}\Leftrightarrow x=\pm\frac{\sqrt{5}}{4}\)
c, \(\frac{2}{\sqrt{x-3}}=4\Leftrightarrow\frac{16}{x-3}=16\)
\(\Leftrightarrow x-3=1\Leftrightarrow x=4\)
d, \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow2x-1=3\Leftrightarrow x=2\)
Câu b tớ thiếu th x âm nhé
c. \(\frac{2}{\sqrt{x-3}}=4\Leftrightarrow2\sqrt{x-3}=1\Leftrightarrow4\left(x-3\right)=1\Leftrightarrow x=\frac{13}{4}\)
d. \(\sqrt{4x^2-4x+1}=3\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
