Question

Giải giúp mình bài 1 bài 2 đi mn

Answer 1

Câu 2: b. \(\sqrt{9x^2-6x+1}=9\)

<=> \(\sqrt{\left(3x-1\right)^2}=9\)

<=> 3x - 1 = 9

<=> 3x = 10

<=> x = \(\dfrac{10}{3}\)

Answer 2

sorry bn, mik chưa hc lớp 9

Answer 3

Câu 2: 

a: \(\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}=6\)

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

\(\Leftrightarrow x+5=4\)

hay x=-1

b: \(\sqrt{9x^2-6x+1}=9\)

\(\Leftrightarrow\left|3x-1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=9\\3x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)