1:
1: \(2\sqrt{12}+3\sqrt{18}-3\sqrt{75}-\sqrt{50}\)
\(=4\sqrt{3}-15\sqrt{3}+9\sqrt{2}-5\sqrt{2}\)
\(=-11\sqrt{3}+4\sqrt{2}\)
2: \(\sqrt{16-6\sqrt{7}}+\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(=\left|3-\sqrt{7}\right|+\left|3+\sqrt{7}\right|\)
\(=3-\sqrt{7}+3+\sqrt{7}=6\)
3:
\(\dfrac{6}{\sqrt{7}-1}+\dfrac{4}{\sqrt{7}-3}+\dfrac{7}{\sqrt{7}}\)
\(=\dfrac{6\left(\sqrt{7}+1\right)}{7-1}-\dfrac{4}{3-\sqrt{7}}+\sqrt{7}\)
\(=\sqrt{7}+1+\sqrt{7}-\dfrac{4\left(3+\sqrt{7}\right)}{2}\)
\(=2\sqrt{7}+1-2\left(3+\sqrt{7}\right)\)
=1-6
=-5
2:
a: ĐKXĐ: x>=5
\(5\sqrt{x-5}+\sqrt{4x-20}-\sqrt{9x-45}=12\)
=>\(5\sqrt{x-5}+2\sqrt{x-5}-3\sqrt{x-5}=12\)
=>\(4\sqrt{x-5}=12\)
=>\(\sqrt{x-5}=3\)
=>x-5=9
=>x=14(nhận)
2:
ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2-20x+25}=1\)
=>\(4x^2-20x+25=1\)
=>(2x-5)2=1
=>\(\left[{}\begin{matrix}2x-5=1\\2x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)