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\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
\(a,=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{5\left(4-\sqrt{7}\right)}{9}\\ =\dfrac{\sqrt{7}-5-3+\sqrt{7}}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{2\sqrt{7}-8}{2}+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\sqrt{7}-4+2\sqrt{7}+4-\dfrac{20-5\sqrt{7}}{9}\\ =\dfrac{27\sqrt{7}-20+5\sqrt{7}}{9}=\dfrac{32\sqrt{7}-20}{9}\)
\(b,=\dfrac{2\left(\sqrt{6}+2\right)}{2}+\dfrac{2\left(\sqrt{6}-2\right)}{2}+\dfrac{5\sqrt{6}}{6}\\ =\sqrt{6}+2+\sqrt{6}-2+\dfrac{5\sqrt{6}}{6}\\ =\dfrac{12\sqrt{6}+5\sqrt{6}}{6}=\dfrac{17\sqrt{6}}{6}\)
\(c,=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\\ =\dfrac{2\sqrt{5}}{5+2\sqrt{6}-5}=\dfrac{2\sqrt{5}}{2\sqrt{6}}=\dfrac{\sqrt{30}}{6}\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
a)
Có:
\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)
Vì \(\sqrt{117}>\sqrt{116}\) nên \(3\sqrt{13}>2\sqrt{29}\)
b)
Có:
\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)
\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)
Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\) nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)
c)
Có:
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)
Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)
d)
Có:
\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)
\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)
lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)
\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)
a: =(2căn 3-8căn 3)(căn 3-1)
=-6căn 3*(căn 3-1)
=-18+6căn 3
b: \(=\dfrac{6-2\sqrt{5}}{\sqrt{5}-3}-\sqrt{5}+2\)
=-2-căn 5+2=-căn 5
c: \(=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
=\(3\sqrt{2a}-3a\cdot\sqrt{2a}\)
\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)