a: Ta có: \(\sqrt{25x-75}-\dfrac{\sqrt{x-3}}{2}+4\cdot\sqrt{\dfrac{9x-27}{16}}=15\)
\(\Leftrightarrow5\sqrt{x-3}-\dfrac{1}{2}\sqrt{x-3}+4\cdot\dfrac{3}{4}\sqrt{x-3}=15\)
\(\Leftrightarrow\sqrt{x-3}\cdot\dfrac{15}{2}=15\)
\(\Leftrightarrow x-3=4\)
hay x=7
b: Ta có: \(\sqrt{x+9}=3-x\)
\(\Leftrightarrow x+9=\left(3-x\right)^2\)
\(\Leftrightarrow x^2-6x+9-x-9=0\)
\(\Leftrightarrow x\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=7\left(loại\right)\end{matrix}\right.\)
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