Question

Câu 1:giải phương trình (tìm x biết):

1) \(5\sqrt{2x}+1=21\)

2)\(\sqrt{9x^{2^{ }}-6x+1}=2\)

3)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)

Answer 1

1) ĐK: x\(\ge0\)

\(5\sqrt{2x}+1=21\Leftrightarrow5\sqrt{2x}=20\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)

Vậy S={8}

2) \(\sqrt{9x^2-6x+1}=2\Leftrightarrow\sqrt{\left(3x\right)^2-2.3x.1+1^2}=2\Leftrightarrow\sqrt{\left(3x-1\right)^2}=2\Leftrightarrow\left|3x-1\right|=2\Leftrightarrow\)\(\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Vậy S={\(-\dfrac{1}{3};3\)}

3) ĐK: x\(\ge0\)

\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\Leftrightarrow\sqrt{3x}=4\sqrt{6}\Leftrightarrow3x=96\Leftrightarrow x=32\left(tm\right)\)

Vậy S={32}

Answer 2

1, ĐKXĐ:x\(\ge0\)
\(5\sqrt{2x}=20\)\(\Leftrightarrow\sqrt{2x}=4\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)(tm)

Answer 3

2,\(\sqrt{9x^2-6x+1}=2\)
\(\Leftrightarrow\sqrt{\left(3x\right)^2+2.3x.1+1}=2\)
\(\Leftrightarrow\sqrt{\left(3x-1\right)^2}=2\)
\(\Leftrightarrow\left|3x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2\\1-3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Answer 4

3,bình phương 2 vế \(\left(2+\sqrt{6}>0\right)\)
\(\Rightarrow10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2\)
\(\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\)
\(\Leftrightarrow\)\(\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
\(\Leftrightarrow x=32\)