\(a,\sqrt{25-10x+x^2}+7=x\)(\(x\ge5\))
\(\Leftrightarrow\sqrt{\left(5-x\right)^2}=7-x\)
\(\Leftrightarrow5-x=7-x\)
\(\Leftrightarrow0x=2\) (vô nghiệm)
Vậy PT đã cho vô nghiệm
\(b,\sqrt{x+6\sqrt{x-9}}=4\left(x\ge9\right)\)
\(\Leftrightarrow\sqrt{x-9+6\sqrt{x-9}+9}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-9}+3\right)^2}=4\)
\(\Leftrightarrow\sqrt{x-9}+3=4\)
\(\Leftrightarrow\sqrt{x-9}=1\)
\(\Leftrightarrow x-9=1\)
\(\Leftrightarrow x=10\left(TM\right)\)
Vậy PT có nghiệm là \(x=10\)