Bài 7:
a: Ta có: \(\sqrt{2x-5}-2\sqrt{3}=0\)
\(\Leftrightarrow2x-5=12\)
hay \(x=\dfrac{17}{2}\)
b: Ta có: \(\sqrt{7-x}+1=x\)
\(\Leftrightarrow7-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1-7+x=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Bài 7:
h: Ta có: \(\sqrt{4x-20}+6\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(\Leftrightarrow2\sqrt{x-5}+2\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow3\sqrt{x-5}=6\)
\(\Leftrightarrow x-5=4\)
hay x=9
i: Ta có: \(\sqrt{x^2-x+\dfrac{1}{4}}=7.5\)
\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{15}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{15}{2}\\x-\dfrac{1}{2}=\dfrac{-15}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)
