Question

Answer 1

a. \(\sqrt{x^2-6x+9}=5\)

\(\sqrt{x^2-2.x.3+3^2}=5\)

\(\sqrt{\left(x-3\right)^2}=5\)

\(x-3=5\)

\(x=8\)

Answer 2

\(b,ĐK:x\ge5\\ PT\Leftrightarrow7\sqrt{x-5}-3\sqrt{x-5}+2\sqrt{x-5}=12\\ \Leftrightarrow6\sqrt{x-5}=12\Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

Answer 3

b. \(7\sqrt{x-5}-\sqrt{9x-45}+\sqrt{4x-20}=12\)      (ĐK: \(x\ge5\))

<=> \(7\sqrt{x-5}-\sqrt{9\left(x-5\right)}+\sqrt{4\left(x-5\right)}=12\)

<=> \(7\sqrt{x-5}-3\sqrt{x-5}+2\sqrt{x-5}=12\)

<=> \(\sqrt{x-5}.\left(7-3+2\right)=12\)

<=> \(6\sqrt{x-5}=12\)

<=> \(\sqrt{x-5}=2\)

<=> x - 5 = 4

<=> x = 9 (TM)

Answer 4

a: Ta có: \(\sqrt{x^2-6x+9}=5\)

\(\Leftrightarrow\left|x-3\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)