Question

Tìm x

a) \(\sqrt{4x}+\sqrt{\frac{x}{4}}+\frac{1}{2}\sqrt{49x}=6\)

b)3x + \(\sqrt{3x-7}\)= 7

Answer 1

\(a,2\sqrt{x}+\frac{1}{2}\sqrt{x}+\frac{7}{2}\sqrt{x}=6 \Leftrightarrow\sqrt{x}\left(2+\frac{1}{2}+\frac{7}{2}\right)=6\Leftrightarrow6\sqrt{x}=6\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)ĐKXĐ: x \(\ge\) 0

b, ĐKXĐ: x \(\ge\) \(\frac{7}{3}\)

3x - 7 + \(\sqrt{3x-7}\)=0

\(\Leftrightarrow\) (\(\sqrt{3x-7}\))(\(\sqrt{3x-7}\)+1)=0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\sqrt{3x-7}=0\\\sqrt{3x-6=0}\end{matrix}\right.\)\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\frac{7}{3}\\x=2\end{matrix}\right.\)(Tm)