Bài 19:
a) Ta có: \(\sqrt{7+\sqrt{2x}}=3+\sqrt{5}\)
\(\Leftrightarrow\sqrt{2x}+7=\left(3+\sqrt{5}\right)^2\)
\(\Leftrightarrow\sqrt{2x}+7=14+6\sqrt{5}\)
\(\Leftrightarrow\sqrt{2x}=7+6\sqrt{5}\)
\(\Leftrightarrow2x=\left(7+6\sqrt{5}\right)^2\)
\(\Leftrightarrow2x=229+84\sqrt{5}\)
hay \(x=\dfrac{229+84\sqrt{5}}{2}\)
b) Ta có: \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\sqrt{3}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
c) Ta có: \(\sqrt{3x^2-4x}=2x-3\)
\(\Leftrightarrow3x^2-4x=4x^2-12x+9\)
\(\Leftrightarrow4x^2-12x+9-3x^2+4x=0\)
\(\Leftrightarrow x^2-8x+9=0\)
\(\Leftrightarrow\left(x-4\right)^2=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=\sqrt{7}\\x-4=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}+4\\x=-\sqrt{7}+4\end{matrix}\right.\)
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