a)Ta có: \(\sqrt{4x-3}-5=x\)
\(\Leftrightarrow\sqrt{4x-3}=x+5\)
ĐKXĐ: \(\left\{{}\begin{matrix}4x-3\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x\ge3\\x\ge-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{4}\\x\ge-5\end{matrix}\right.\Leftrightarrow x\ge\frac{3}{4}\)
Ta có: \(\sqrt{4x-3}=x+5\)
\(\Leftrightarrow\left(\sqrt{4x-3}\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow4x-3=x^2+10x+25\)
\(\Leftrightarrow x^2+10x+25-4x+3=0\)
\(\Leftrightarrow x^2+6x+28=0\)(Vô lý)
Vậy: S=∅
b) ĐKXĐ: \(\left\{{}\begin{matrix}10-x\ge0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le10\\x\ge-3\end{matrix}\right.\Leftrightarrow-3\le x\le10\)
Ta có: \(\sqrt{10-x}-\sqrt{x+3}=5\)
\(\Leftrightarrow\left(\sqrt{10-x}-\sqrt{x+3}\right)^2=5^2\)
\(\Leftrightarrow10-x+x+3-2\sqrt{\left(10-x\right)\left(x+3\right)}=25\)
\(\Leftrightarrow13-2\sqrt{\left(10-x\right)\left(x+3\right)}=25\)
\(\Leftrightarrow-2\sqrt{\left(10-x\right)\left(x+3\right)}=12\)
\(\Leftrightarrow\sqrt{\left(10-x\right)\left(x+3\right)}=-6\)(vô lý)
Vậy: S=∅
