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g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{3}\)
b) \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}\)
\(=0\)
c) \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\cdot\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)
\(=\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)\)
\(=4-4\sqrt{5}+5\)
\(=9-4\sqrt{5}\)
d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=6-121\)
\(=-115\)
a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
a) \(A=\left(1-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+2\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{\sqrt{3}-1}{2}\right):\left(\dfrac{\sqrt{3}-1}{2}+\dfrac{4}{2}\right)\)
\(=\dfrac{2-\left(\sqrt{3}-1\right)}{2}:\dfrac{\left(\sqrt{3}-1\right)+4}{2}\)
\(=\dfrac{3-\sqrt{3}}{2}.\dfrac{2}{\sqrt{3}+3}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2}{2}\)
Vì \(\left\{{}\begin{matrix}\left(\sqrt{3}-1\right)^2>0\\2>0\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(\sqrt{3}-1\right)^2}{2}>0\) hay A>0
=> A có căn bậc 2
Vậy......
b)\(B=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\sqrt{5}\right):\dfrac{\sqrt{5}+\sqrt{2}}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\left(\dfrac{\sqrt{2}\left(3-1\right)}{1-3}-\sqrt{5}\right).\dfrac{5-2}{\sqrt{5}+\sqrt{2}}\)
\(=\left(-\sqrt{2}-\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right).\dfrac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
Vì -3 < 0 hay B < 0
=> B không có căn bậc 2
Vậy.....
