Question

[yung Shin]

Rút gọn bt

a) \(\dfrac{3-\sqrt{x}}{x-9}\:vớix\ge0,x\ne9\)

b) 6-2x-\(\sqrt{9-6x+x^2}vớix< 3\)

tìm x biết

a)\(\sqrt{1-12x+36x^2}=5\)

Answer 1

[Nguyễn Huy Tú]

Bài 1:

a, \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-1}{\sqrt{x}+3}\)

b, \(6-2x-\sqrt{9-6x+x^2}\)

\(=6-2x-\sqrt{\left(3-x\right)^2}\)

\(=6-2x-3+x\left(x< 3\right)\)

\(=3-x\)

Bài 2:

\(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\sqrt{\left(1-6x\right)^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

+) Xét \(x\ge\dfrac{1}{6}\) có:
\(6x-1=5\Leftrightarrow x=1\)

+) Xét \(x< \dfrac{1}{6}\) có:

\(1-6x=5\)

\(\Leftrightarrow x=\dfrac{-2}{3}\)

Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\end{matrix}\right.\)