Question

3/ Tìm x,biết:

a) 3 √𝑥−3=12

b) √16(1−2𝑥)−8=0

c) √4(9−6𝑥+𝑥2)−12= 0

Answer 1

a) \(3\sqrt{x-3}=12\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x-3}=4\)

\(\Leftrightarrow x-3=16\Leftrightarrow x=19\left(tm\right)\)

b) \(\sqrt{16\left(1-2x\right)}-8=0\left(đk:x\le\dfrac{1}{2}\right)\)

\(\Leftrightarrow4\sqrt{1-2x}=8\Leftrightarrow\sqrt{1-2x}=2\)

\(\Leftrightarrow1-2x=4\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)

c) \(\sqrt{4\left(9-6x+x^2\right)}-12=0\)

\(\Leftrightarrow2\sqrt{\left(x-3\right)^2}=12\)

\(\Leftrightarrow\left|x-3\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=6\\x-3=-6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)

Answer 2

a: ta có: \(3\sqrt{x-3}=12\)

\(\Leftrightarrow x-3=16\)

hay x=19

b: Ta có: \(\sqrt{16\left(1-2x\right)}-8=0\)

\(\Leftrightarrow1-2x=4\)

\(\Leftrightarrow2x=-3\)

hay \(x=-\dfrac{3}{2}\)