a, \(\forall x=>\sqrt{x^2-10x+25}=7\)
\(< =>\sqrt{\left(x-5\right)^2}=7< =>|x-5|=7\)
\(= >\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)
b, đK; \(x\ge\dfrac{3}{5}=>\sqrt{25x^2-9}=2\sqrt{5x-3}\)
\(< =>\sqrt{5x-3}\left(\sqrt{5x+3}-2\right)=0\)
\(=>\left[{}\begin{matrix}\sqrt{5x-3}=0\\\sqrt{5x+3}-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)
c,đK" \(x\ge3=>\dfrac{\sqrt{x-3}}{\sqrt{2x+1}}=2\)
\(=>\sqrt{x-3}=2\sqrt{2x+1}< =>x-3=4\left(2x+1\right)\)
\(< =>x=-1\left(ktm\right)\)
d, đk: \(x\ge5\)
=>VT\(=\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{5}.3\sqrt{x-5}=3\)
\(< =>\dfrac{12}{5}\sqrt{x-5}=3=>\sqrt{x-5}=\dfrac{5}{4}=>x-5=\dfrac{25}{16}=>x=\dfrac{105}{16}\left(tm\right)\)
