Bài 3:
a: \(A=2a^2b\cdot\sqrt{\frac{64}{a^4b^2}}\)
\(=2a^2b\cdot\frac{8}{a^2\left|b\right|}\)
\(=\frac{16b}{b}\left(b>0\right)\)
=16
b:
\(a>\frac12\)
=>2a>1
=>2a-1>0
\(B=\sqrt{\frac{27}{48\left(4a^2-4a+1\right)}}\)
\(=\sqrt{\frac{9}{16\left(2a-1\right)^2}}=\frac{3}{4\left|2a-1\right|}\)
\(=\frac{3}{4\left(2a-1\right)}\)
c: a>b>0 nên a-b>0
\(C=\left(a-b\right)\cdot\sqrt{\frac{a^6b^2}{a^2-2ab+b^2}}\)
\(=\left(a-b\right)\cdot\frac{\sqrt{a^6b^2}}{\sqrt{\left(a-b\right)^2}}\)
\(=\left(a-b\right)\cdot\frac{a^3b}{a-b}=a^3b\)
d: 4a>b
=>\(\sqrt{4a}>\sqrt{b}\)
=>\(2\sqrt{a}>\sqrt{b}\)
=>\(2\sqrt{a}-\sqrt{b}>0\)
\(D=\frac{\sqrt{16a^2-8ab+b^2}}{10\left(2a-\sqrt{ab}\right)}\)
\(=\frac{\sqrt{\left(4a-b\right)^2}}{10\cdot\sqrt{a}\cdot\left(2\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4a-b}{10\sqrt{a}\left(2\sqrt{a}-\sqrt{b}\right)}=\frac{\left(2\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}+\sqrt{b}\right)}{10\sqrt{a}\left(2\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{2\sqrt{a}+\sqrt{b}}{10\sqrt{a}}\)
Bài 2:
a: \(A=2\sqrt2\left(5+2\sqrt8-3\sqrt{18}\right)\)
\(=2\sqrt2\left(5+2\cdot2\sqrt2-3\cdot3\sqrt2\right)\)
\(=2\sqrt2\left(5+4\sqrt2-9\sqrt2\right)=2\sqrt2\left(5-5\sqrt2\right)=10\sqrt2-20\)
b: \(B=\sqrt{13+6\cdot\sqrt{4+\sqrt{9-4\sqrt2}}}\)
\(=\sqrt{13+6\cdot\sqrt{4+\sqrt{\left(2\sqrt2-1\right)^2}}}\)
\(=\sqrt{13+6\cdot\sqrt{4+\left(2\sqrt2-1\right)}}\)
\(=\sqrt{13+6\cdot\sqrt{3+2\sqrt2}}\)
\(=\sqrt{13+6\cdot\sqrt{\left(\sqrt2+1\right)^2}}=\sqrt{13+6\left(\sqrt2+1\right)}=\sqrt{19+6\sqrt2}\)
\(=\sqrt{\left(3\sqrt2+1\right)^2}=3\sqrt2+1\)
c: \(C=\left(\sqrt3-1\right)\cdot\sqrt{2+\sqrt{19+8\sqrt3}-4}\)
\(=\left(\sqrt3-1\right)\cdot\sqrt{2+\sqrt{\left(4+\sqrt3\right)^2}-4}\)
\(=\left(\sqrt3-1\right)\cdot\sqrt{4+\sqrt3-2}=\left(\sqrt3-1\right)\cdot\sqrt{2+\sqrt3}\)
\(=\left(\sqrt3-1\right)\cdot\frac{\sqrt{4+2\sqrt3}}{\sqrt2}\)
\(=\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)}{\sqrt2}=\frac{3-1}{\sqrt2}=\frac{2}{\sqrt2}=\sqrt2\)
d: \(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt6\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\cdot\sqrt{\left(\sqrt5-\sqrt3\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt5-\sqrt3\right)\left(\sqrt5-\sqrt3\right)\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=32-8\sqrt{15}+8\sqrt{15}-30\)
=2
