Question

Tìm x, biết:

1)\(3\left(\sqrt{x}+2\right)+5=4\sqrt{4x}+1\)

2) \(\sqrt{1-3x}< 2\)

Answer 1

1) ĐKXĐ: x≥0

Ta có: \(3\left(\sqrt{x}+2\right)+5=4\sqrt{4x}+1\)

\(\Leftrightarrow3\sqrt{x}+6+5=8\sqrt{x}+1\)

\(\Leftrightarrow3\sqrt{x}-8\sqrt{x}=1-6-5\)

\(\Leftrightarrow-5\sqrt{x}=-10\)

\(\Leftrightarrow5\sqrt{x}=10\)

\(\Leftrightarrow\sqrt{x}=2\)

hay x=4(nhận)

Vậy: S={4}

2) ĐKXĐ: \(x\le\frac{1}{3}\)

Ta có: \(\sqrt{1-3x}< 2\)

\(\Leftrightarrow\left|1-3x\right|< 4\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x< 4\\1-3x>-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x< 3\\-3x>-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x< \frac{5}{3}\end{matrix}\right.\)

Kết hợp ĐKXĐ, ta được: \(-1< x\le\frac{1}{3}\)