b) \(\sqrt{x^2-10x+25}=7-2x\)
<=> \(\sqrt{\left(x-5\right)^2}=7-2x\)
<=> \(x-5=7-2x\)
<=> 3x = 12 <=> x = 4
a/ đk: x ≥ 0
\(x-9\sqrt{x}+14=0\)
\(\Leftrightarrow x-2\sqrt{x}-7\sqrt{x}+14=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-7\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-7=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=49\end{matrix}\right.\)
b/ \(\sqrt{x^2-10x+25}=7-2x\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\)
\(\Leftrightarrow\left|x-5\right|=7-2x\)
+) Với x - 5 ≥ 0 <=> x ≥ 5 ta có:
x - 5 = 7 - 2x
<=> 3x = 12 <=> x = 4 (không t/m)
+) Với x - 5 < 0 <=> x < 5 ta có:
x - 5 = 2x - 7
<=> -x = -2
<=> x = 2 (t/m)
Vậy pt có nghiệm x = 2
p/s: ý b bạn kia làm sai nên mk làm lại
