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<=> A=\(\left(\sqrt{\sqrt{7+4\sqrt{3}}}-\sqrt{\sqrt{16-2.4.2\sqrt{3}+\left(2\sqrt{3}\right)^2}}\right)\sqrt{\sqrt{7+4\sqrt{3}}}\)
= \(\left(\sqrt{\sqrt{\left(2+\sqrt{3}\right)^2}}-\sqrt{\sqrt{\left(4-2\sqrt{3}\right)^2}}\right)\sqrt{\sqrt{\left(2+\sqrt{3}\right)^2}}\)
=\(\left(\sqrt{2+\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\sqrt{2+\sqrt{3}}\)
= \(2+\sqrt{3}-\sqrt{\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
=\(2+\sqrt{3}-\sqrt{2\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
=\(2+\sqrt{3}-\sqrt{2\left(4-3\right)}\)
=\(2+\sqrt{3}-\sqrt{2}\)
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(M>0\Leftrightarrow M^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}..\)
\(M^2=8-2.\sqrt{16-7}=8-6=3\)
\(M=\sqrt{3}.\)
a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)
\(a.\sqrt{14+4\sqrt{3}.\sqrt{2}}=\sqrt{12+2.2\sqrt{3}.\sqrt{2}+2}=2\sqrt{3}+\sqrt{2}\)
\(b.\sqrt{11-4\sqrt{3}.\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.\sqrt{3}+3}=2\sqrt{2}-\sqrt{3}\)
\(c.\sqrt{28+16\sqrt{3}}=\sqrt{16+2.2\sqrt{3}.4+12}=4+2\sqrt{3}\)
\(d.\sqrt{11+4\sqrt{7}}=\sqrt{7+2.2\sqrt{7}+4}=\sqrt{7}+2\)
\(e.\sqrt{29-4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}+1}=2\sqrt{7}-1\)
\(f.\sqrt{21+6\sqrt{2}.\sqrt{3}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}\)
a: \(B=3\sqrt{7}-6\sqrt{12}+\sqrt{84}-24\)
b: \(B=3\cdot\dfrac{\sqrt{2}-\sqrt{5}}{\left(\sqrt{2}-\sqrt{5}\right)^2}=\dfrac{3}{\sqrt{2}-\sqrt{5}}=-\sqrt{5}-\sqrt{2}\)
Lời giải :
a) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)
\(=0,1-\sqrt{0,1}\)
b) \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
c) \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
d) \(\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)
e) \(\sqrt{16-6\sqrt{7}}=\sqrt{9-2\cdot3\cdot\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7}\)