Bài 1:
a)
\(A=\sqrt{2}\left(\sqrt{4.2}-\sqrt{16.2}+3\sqrt{9.2}\right)\\ =\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\\ =\sqrt{2}.7\sqrt{2}\\ =7\)
b)
\(B=\dfrac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{6-1}+\dfrac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{2-3}+\sqrt{\sqrt{2}^2-2.\sqrt{2}.\sqrt{1}+\sqrt{1}^2}\\ =\dfrac{5\left(\sqrt{6}-1\right)^2}{5}-\left(\sqrt{2}-\sqrt{3}\right)^2+\sqrt{\left(\sqrt{2}-1\right)^2}\\ =5\left(6-2\sqrt{6}+1\right)-\left(2-2\sqrt{6}+3\right)+\sqrt{2}-1\\ =30-10\sqrt{6}+5-2+2\sqrt{6}-3+\sqrt{2}-1\\ =29-8\sqrt{6}\)
2:
a: \(\sqrt{x^2-2x+9}=x+2\)
=>x>=-2 và x^2-2x+9=x^2+4x+4
=>x>=-2 và -2x+9=4x+4
=>x>=-2 và -6x=-5
=>x=5/6(nhận)
b:
ĐKXĐ: x^2-4>=0 và x+2>=0
=>x>=-2 và (x>=2 hoặc x<=-2)
=>x=-2 hoặc x>=2
\(\sqrt{x^2-4}+\sqrt{x+2}=0\)
=>x^2-4=0 và x+2=0
=>x=-2
c:
ĐKXĐ: x>=1
\(\sqrt{x+3-4\sqrt{x-1}}=3\)
=>\(\sqrt{x-1-2\cdot\sqrt{x-1}\cdot2+4}=3\)
=>\(\left|\sqrt{x-1}-2\right|=3\)
=>\(\left[{}\begin{matrix}\sqrt{x-1}-2=3\\\sqrt{x-1}-2=-3\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=5\)
=>x-1=25
=>x=26
d: \(\sqrt{4-8x}-4\sqrt{1-2x}+\sqrt{\dfrac{9-18x}{4}}+1=0\)
=>\(2\sqrt{1-2x}-4\sqrt{1-2x}+\dfrac{3}{2}\sqrt{1-2x}+1=0\)
=>\(1-\dfrac{1}{2}\sqrt{1-2x}=0\)
=>\(\sqrt{1-2x}=2\)
=>1-2x=4
=>2x=-3
=>x=-3/2
