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a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)
\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)
\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)
\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)
\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)
\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)
\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
a)
=\(\sqrt{18-2.3\sqrt{2}.1+1}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1\)
b)
=\(\sqrt{12+2.2\sqrt{3}.3+9}\)
=\(\sqrt{\left(2\sqrt{3}+3\right)^2}\)
=\(2\sqrt{3}+3\)
c)
=\(\sqrt{25-2.5.4\sqrt{2}+32}\)
=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)
=\(4\sqrt{2}-5\)
d)
\(=\sqrt{\left(3-2.\sqrt{3}.\sqrt{2}+2\right)\left(3-2\sqrt{3}+1\right)}\\ =\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}\\ =\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\\ =3-\sqrt{3}-\sqrt{6}+\sqrt{2}\)
e)
\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)
g)
\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
c) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\) = \(\dfrac{\sqrt{12-8\sqrt{2}}}{\sqrt{2}}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
= \(\dfrac{\sqrt{\left(2\sqrt{2}-2\right)^2}}{\sqrt{2}}+\sqrt{\left(3\sqrt{2}-1\right)^2}\) = \(\dfrac{2\sqrt{2}-2}{\sqrt{2}}+3\sqrt{2}-1\)
\(\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{\sqrt{2}}+3\sqrt{2}-1\) = \(2-\sqrt{2}+3\sqrt{2}-1\) = \(2\sqrt{2}+1\)
d )Đặt A = \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(\Leftrightarrow A^2=\left(\sqrt{12-3\sqrt{7}}\right)^2-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}+\left(\sqrt{12+3\sqrt{7}}\right)^2\)
\(\Leftrightarrow A^2=12-3\sqrt{7}-2\sqrt{144-63}+12+3\sqrt{7}\)
\(\Leftrightarrow A^2=24-2\sqrt{81}\)
\(\Leftrightarrow A^2=24-18=6\)
=> A = \(\sqrt{6}\)
Vậy \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=\sqrt{6}\)